Algebra 3: algorithms in algebra
Hans Sterk
2003-2004
ii
Contents
1 Polynomials, Gröbner bases and Buchberger’s algorithm
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Polynomial rings and systems of polynomial equations . . .
1.3 Monomial orderings . . . . . . . . . . . . . . . . . . . . . .
1.4 A division algorithm for polynomials . . . . . . . . . . . . .
1.5 Monomial ideals and Gröbner bases . . . . . . . . . . . . .
1.6 Buchberger’s algorithm . . . . . . . . . . . . . . . . . . . . .
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2 Applications
2.1 Elimination . . . . . . . . . . . . . . . .
2.2 Geometry theorem proving: first glimpse
2.3 The Nullenstellensatz . . . . . . . . . .
2.4 Algebraic numbers . . . . . . . . . . . .
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3 Factorisation of polynomials
3.1 Introduction . . . . . . . . . . . . . . . .
3.2 Polynomials with integer coefficients . .
3.3 Factoring polynomials modulo a prime .
3.4 Factoring polynomials over the integers
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4 Symbolic integration
4.1 Introduction . . . . . . . .
4.2 Differential fields . . . . .
4.3 Rational functions . . . .
4.4 Beyond rational functions
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A Algebraic prerequisites
53
A.1 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
A.2 Rings, ideals and quotient rings . . . . . . . . . . . . . . . . . 54
A.3 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
i
CONTENTS
A.4 Resultants . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.5 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
57
59
2
CONTENTS
Chapter 1
Polynomials, Gröbner bases
and Buchberger’s algorithm
1.1
Introduction
This chapter deals with the algebraic approach to systems of polynomial
equations. Rather than manipulating polynomial equations directly, this
approach focuses on studying ideals in polynomial rings and finding generators for these ideals suitable for various types of computations. The suitable
sets of generators we are looking for are the so–called Gröbner bases (introduced in Section 1.5). In Section 1.6 we discuss Buchberger’s algorithm
to construct such Gröbner bases. The algorithm can be seen as a common
generalization of the Euclidean algorithm for gcd computations (for polynomials in one variable) and the Gauss–Jordan procedure for solving systems
of linear equations.
To analyse ideals we need a bit of the machinery of rings in the context
of polynomial rings, and, most significantly, an ordering on the set of monomials in polynomial rings to enable us to generalize division with remainder
to polynomials in several variables. These items are explained in Sections
1.2, 1.3 and 1.4.
1.2
Polynomial rings and systems of polynomial
equations
1.2.1 Instead of considering a system of poynomial equations
f1 = 0, f2 = 0, . . . , fm = 0,
1
2
Polynomials, Gröbner bases and Buchberger’s algorithm
it turns out to be more fruitful to study the ideal (f1 , . . . , fm ) in the corresponding polynomial ring. This section introduces the terminology and
notations regarding polynomial rings over a field. Let k[X1 , . . . , Xn ] be a
polynomial ring in n indeterminates over the field k.
1.2.2 Definition. A monomial is an element of k[X1 , . . . , Xn ] of the form
X1m1 X2m2 · · · Xnmn .
The (multi)degree of this monomial is the vector m = (m1 , . . . , mn ); the
total degree is the sum m1 + · · · + mn and is often denoted by |m|. If n = 1,
the notions coincide with the usual notion of degree. We often shorten the
notation by writing X m for X1m1 X2m2 · · · Xnmn .
1.2.3
Every polynomial is a finite k–linear combination of monomials.
The first result states that every ideal in a polynomial ring is finitely generated. This is a consequence of a more general result, which we derive in a
moment.
1.2.4 Definition. A ring R is called noetherian if every ideal in R is finitely
generated.
Here is an equivalent way of phrasing the property.
1.2.5 Lemma. A ring R is noetherian if and only if every ascending chain of
ideals I1 ⊂ I2 ⊂ I3 ⊂ · · · in R stabilizes (i.e., there is an index m such that
Im = Im+1 = Im+2 = · · · ).
1.2.6 Theorem. (Hilbert basis theorem) If R is noetherian, then so is R[X].
Proof. Suppose I ⊂ R[X] is not finitely generated. Choose f1 ∈ I \ {0} of
minimal degree d1 . Then choose f2 ∈ I \ (f1 ) of minimal degree n2 (this is
possible since f1 cannot generate I by assumption), etc. Now suppose fi =
adi X di + · · · and consider the chain of ideals (a1 ) ⊂ (a1 , a2 ) ⊂ (a1 , a2 , a3 ) ⊂
· · · in R. This chain stabilizes at some point by the lemma. Let’s say that
ak+1 ∈ (a1 , . . . , ak ) and write ak+1 = b1 a1 +· · ·+bk ak for some b1 , . . . , bk ∈ R.
Then the polynomial
g := fk+1 −
k
X
bi fi X dk+1 −di
i=1
has, by construction, degree less than the degree of fk+1 , but is, like fk+1 ,
not an element of (f1 , . . . , fk ). This is a contradiction.
¤
1.3 Monomial orderings
3
1.2.7 Since a field k has only two ideals, (0) and k itself, every field is a noetherian
ring. By applying the Hilbert basis theorem several times we find that the
polynomial ring in n indeterminates over a field is noetherian. Every ideal
in such a ring is therefore finitely generated.
1.2.8 Corollary. Every polynomial ring over a field is noetherian. If I is an
ideal in such a ring then there exist elements f1 , . . . , fs ∈ I such that I =
(f1 , f2 , . . . , fn ).
1.2.9 The importance of this result is that every system of polynomial equations
in n variables can be replaced by an equivalent finite system. With this
observation, the problem of finding the zeros of a system of polynomial
equations is equivalent to the problem of finding the common zeros of an
ideal. If I is an ideal in k[X1 , . . . , Xn ], then we define V (I), the zeroset of I
as
V (I) = {(a1 , . . . , an ) ∈ k n | f (a1 , . . . , an ) = 0 for all f ∈ I}.
If I = (f1 , . . . , fs ), then it is easy to see that
V (I) = {(a1 , . . . , an ) ∈ k n | fi (a1 , . . . , an ) = 0, i = 1, . . . , s}.
In particular, if I = (f1 , . . . , fs ) = (g1 , . . . , gt ), then the systems of equations
f1 = 0, . . . , fs = 0
and
g1 = 0, . . . , gt = 0
are equivalent. With the appearance of ideals, the whole machinery of commutative algebra is at our disposal to analyse their structure and properties.
1.3
Monomial orderings
1.3.1 In this section we discuss various ways to order the monomials of a polynomial ring. This is needed in order to set up a division algorithm. This
algorithm imitates the one for polynomials in one variable. For polynomials
in one variable X, there is one ordering: 1 < X < X 2 < X 3 < · · · that
makes sense for division purposes. In the case of several variables, there are
more possibilities.
1.3.2 Definition. A partial order on a set S is a relation ≥ on S such that
4
Polynomials, Gröbner bases and Buchberger’s algorithm
(i) a ≥ a for every a ∈ S (the relation is reflexive);
(ii) if a ≥ b and b ≥ c then a ≥ c (the relation is transitive);
(iii) a ≥ b and b ≥ a imply a = b (the relation is antisymmetric).
A partial order is called a total order if, in addition,
(iv) for all a, b ∈ S, either a ≤ b or b ≤ a.
A total order is called a well–ordering if moreover the following holds:
(v) Every nonempty subset T of S contains a smallest element: there is a
t ∈ T such that t ≤ s for all t ∈ T .
1.3.3 Remark. In dealing with monomials X a = X1a1 · · · Xnan , we will often just
use the exponent vector a = (a1 , . . . , an ) ∈ Zn≥0 instead of the whole monomial.
1.3.4 Definition. (Monomial ordering) A monomial ordering on k[X1 , . . . , Xn ]
is a well–ordering on the set of monomials X a such that
a > b and c ∈ Zn≥0
⇒
a + c > b + c.
The condition means that the ordering behaves well with respect to multiplication by monomials. In the following we will define three orderings.
1.3.5 Definition. (Lexicographic order) a >lex b (or X a >lex X b ) if the first
nonzero entry from the left in a − b is positive. We often abbreviate lexicographic order to ‘lex order’.
1.3.6 Definition. (Graded lex order) a >grlex b (or X a >grlex X b ) if |a| > |b|
or |a| = |b| and a >lex b. Graded lex order orders by total degree first and
breaks ties using lex order.
1.3.7 Definition. (Graded reverse lex order) a >grevlex b (or X a >grevlex
X b ) if |a| > |b| or |a| = |b| and the first nonzero entry from the right
in a − b is negative.
1.3.8 Proposition. The lexicographic order, graded lex order and graded reverse
lex order are monomial orderings.
1.4 A division algorithm for polynomials
5
Proof. (Sketch for lex order) Most of the conditions to be verified are
straightforward. To check that lex order is a well–ordering we use the observation that a total order on Zn≥0 is a well–ordering if and only if every
decreasing sequence a(1) > a(2) > · · · terminates. To check this condition
for the lex order, one first considers the first (from the left) coordinates in
such a sequence a(1) > a(2) > · · · ; since they all come from Z≥0 , they
become constant from some point onwards. Next consider from that point
on the second coordinates, etc.
¤
1.3.9 Now that we have the notion of an ordering on monomials, we can refine
our definitions regarding the terms in a polynomial.
P
1.3.10 Definition. If f = a ca X a is a polynomial in k[X1 , . . . , Xn ] and > is a
monomial ordering, then we define
• the multidegree of f to be the maximum degree of the nonzero terms
of f ;
• the leading term lt(f ) of f to be the nonzero term ca X a of f of maximum degree and the leading monomial to be the monomial X a ;
• the leading coefficient lc(f ) of f to be the coefficient of the leading
term of f .
1.4
A division algorithm for polynomials
1.4.1 Given a monomial ordering on a polynomial ring, we can mimic the division
algorithm for polynomials in one variable. However, in the general case
some complications arise when doing a division. Before we turn to the
division algorithm, we remark that division is related to the question of
deciding whether a given element belongs to an ideal. In the case of one
variable this is clear: if an ideal is generated by f1 , . . . , fs , say, then it is
also generated by the gcd (greatest common divisor) f of these elements.
To decide if an element h belongs to the ideal, perform a division with
remainder: h = qf + r. Then h is in the ideal if and only if r = 0.
Fix a monomial ordering on the polynomial ring k[X1 , . . . , Xn ]. We will
describe how to divide a given polynomial f by the polynomials f1 , . . . , fs .
The result will consist of a list of s ‘quotients’ q1 , . . . , qs and a ‘remainder’ r.
These polynomials will be constructed along the way. First they are all set
to 0. In each stage the polynomial f will change: this changing polynomial
is called p; at the beginning it is set equal to f .
6
Polynomials, Gröbner bases and Buchberger’s algorithm
In each stage the algorithm works roughly as follows.
• Look for the first polynomial among the fi (starting from f1 ) whose
leading term divides the leading term of p. If such a division occurs
for fi , then subtract
lt(p)
fi
lt(fi )
from p and add
lt(p)
lt(fi )
to the i–th quotient qi .
• If for no i division occurs, then subtract the term lt(p) from p and add
this term to the remainder r.
Since in each step the leading term of p decreases, this process must terminate. Upon termination, we have an equality
f = q1 f1 + · · · + qs fs + r,
where r = 0 or no term of r is divisible by any of the leading terms of the
fi (i = 1, . . . , s). The remainder and the quotients need not be unique (as
in the one variable case). In fact, the results even depend in general on the
order of the fi . We will come back to these matters soon.
1.5
Monomial ideals and Gröbner bases
In this section we assume that a monomial ordering is specified on the polynomial ring k[X1 , . . . , Xn ]. Before we can state the definition of a Gröbner
basis, we need some preparations involving monomials.
1.5.1 Definition. (Monomial ideals) A monomial ideal in k[X1 , . . . , Xn ] is an
ideal generated by monomials. Note that it is harmless to replace ‘monomial’
by ‘term’, since monomials and terms differ by a constant.
1.5.2 Lemma. Let I be a monomial ideal generated by the monomials X a , a ∈ A.
(a) f ∈ I ⇔ every term/monomial of f is in I.
(b) X b ∈ I ⇔ X a | X b for some a ∈ A.
(c) I = (X a(1) , X a(2) , . . . , X a(m) ) for some m.
1.5 Monomial ideals and Gröbner bases
7
Proof. (a) and (b) follow from writing out an expressions for f and X b ,
respectively. Item (c): Apply the Hilbert basis theorem to get finitely many
(possibly non–monomial) generators. Then apply the previous items to replace these generators by finitely many monomials from the original generating set.
¤
1.5.3 Definition. For an ideal (0) 6= I ⊂ k[X1 , . . . , Xn ] we let lt(I) be the set of
leading terms in I:
lt(I) = {lt(f ) | 0 6= f ∈ I}.
The leading term ideal is the ideal (lt(I)) generated by lt(I).
1.5.4 Lemma. Let (0) 6= I ⊂ k[X1 , . . . , Xn ] be an ideal, then (lt(I)) is a monomial ideal and there exist finitely many elements f1 , . . . , fs ∈ I such that
lt(f1 ), . . . , lt(fs ) generate this ideal.
Proof. This follows from the definition and the previous lemma.
¤
1.5.5 Definition. (Gröbner basis) A finite subset {g1 , . . . , gs } of the ideal I is
called a Gröbner basis of I if the leading term ideal is generated by the
leading terms of the gi :
(lt(g1 ), . . . , lt(gs )) = (lt(I)).
Here is the first important result about Gröbner bases.
1.5.6 Theorem. Let I 6= (0) be an ideal in the polynomial ring k[X1 , . . . , Xn ].
(a) The ideal I has a Gröbner basis.
(b) A Gröbner basis {g1 , . . . , gs } of I generates I (as an ideal):
(g1 , . . . , gs ) = I.
(c) If {g1 , . . . , gs } is a Gröbner basis for I, then division by g1 , . . . , gs leaves
a unique remainder r independent of the order of the gi . In fact, r is
characterized as the unique polynomial such that
(i) r = 0 or no term of r is divisible by any of the leading terms of
the gi (i = 1, . . . , s);
(ii) f − r ∈ I.
8
Polynomials, Gröbner bases and Buchberger’s algorithm
Proof. (a) follows from Lemma 1.5.2 since the leading term ideal (lt(I)) of
I is generated by the leading terms of elements 6= 0 of I.
(b) It is obvious that (g1 , . . . , gs ) ⊂ I as all the gj are in I and therefore
also the ideal generated by them. For the converse inclusion let f ∈ I
and use division with remainder to write f = q1 g1 + · · · + qs gs + r, where
either r = 0 or no term of r is divisible by any of the leading terms lt(gj )
(j = 1, . . . , s). Suppose r 6= 0. From r = f − (q1 g1 + · · · + qs gs ) we conclude
that r ∈ I and so lt(r) ∈ (lt(I)). Again by Lemma 1.5.2 we find that lt(r)
is divisible by one of the leading terms lt(gj ) (j = 1, . . . , s), a contradiction.
So r must be 0 and f ∈ (g1 , . . . , gs ).
(c) Division with remainder as discussed before shows that the remainder
r satisfies the two properties mentioned. So existence of such an r is clear.
As for uniqueness, if r̃ also satisfies the two properties, then r − r̃ ∈ I
(subtract f − r̃ and f − r) and so the leading term of r − r̃ belongs to (lt(I)),
if r − r̃ 6= 0. In this case, as in (b), we arrive at a contradiction, because no
term of r − r̃ is divisible by any of the lt(gj ), wheras Lemma 1.5.2 implies
that one of them is. So r − r̃ = 0, i.e., r = r̃.
¤
1.6
Buchberger’s algorithm
Apart from the definition given in the previous section, there are various
ways of characterizing Gröbner bases. Some of these characterizations together with further properties of Gröbner bases lead to an algorithmic approach to computing Gröbner bases: Buchberger’s algorithm.
We begin with an application of the division algorithm. It settles the
‘ideal membership problem’.
1.6.1 Proposition. (Ideal membership test) Let G be a Gröbner basis for the
ideal I ⊂ k[X1 , . . . , Xn ] and let f ∈ k[X1 , . . . , Xn ]. Then f ∈ I if and only
if the remainder on division of f by G is zero.
Proof. The implication ‘If’ is trivial. Conversely, if f ∈ I, then f = f + 0
satisfies the properties of Theorem 1.5.6, so 0 is the remainder upon division.
¤
1.6.2
In a given Gröbner basis there may be elements of redundancy. For example, if G = {g1 , . . . , gs } is a Gröbner basis for I and if lt(f ) is contained in
the ideal (lt(G − {f }) for f ∈ G, then G − {f } is also a Gröbner basis for
I. Given the definition of Gröbner basis, this is almost a triviality: Since
1.6 Buchberger’s algorithm
9
lt(f ) ∈ (lt(G−{f }), we find (lt(G−{f }) = (lt(G)) = (lt(I)). The resulting
equality of the first and third term imply that G − {f } is a Gröbner basis.
The following definitions and results are intended to produce ‘unique’
Gröbner bases in some sense.
1.6.3 Definition. A minimal Gröbner basis for an ideal I is a Gröbner basis G
for I satisfying
(i) lc(f ) = 1 for all f ∈ G;
(ii) lt(f ) 6∈ (lt(G − {f })) for all f ∈ G.
A reduced Gröbner basis for an ideal I satisfies (i) and the following condition, which is stronger than (ii):
(ii’) no (nonzero) term of f is in (lt(G − {f })) for all f ∈ G.
1.6.4 Theorem. Every nonzero ideal I ⊂ k[X1 , . . . , Xn ] has a unique reduced
Gröbner basis (for a given monomial ordering).
Proof. Given a Gröbner basis, it is easy to construct a minimal one: just apply the observation we made above in 1.6.2 and replace leading coefficients.
To construct a reduced one is less obvious.
¤
1.6.5 (Equality of ideals) Once reduced Gröbner bases can be effectively computed, one has a method to decide whether two ideals are equal: they are
equal if and only if they the same reduced Gröbner basis.
1.6.6 Definition. For nonzero polynomials f, g ∈ k[X1 , . . . , Xn ] of multidegree a
and b, respectively, we define their S–polynomial as the polynomial
S(f, g) =
Xc
Xc
·f −
· g,
lt(f )
lt(g)
where c = (max(a1 , b1 ), . . . , max(an , bn )). The monomial X c is called the
least common multiple (lcm) of the leading terms of f and g.
1.6.7 S–polynomials are vital ingredients in Buchberger’s algorithm for computing Gröbner bases. Here is a characterization of Gröbner bases involving
S–polynomials.
1.6.8 Theorem. A basis G = {g1 , . . . , gs } for the nonzero ideal I is a Gröbner
basis for I if and only if the remainder on division of S(gi , gj ) by G is zero
for all i 6= j.
10
Polynomials, Gröbner bases and Buchberger’s algorithm
Proof. The proof of this theorem is not deep, but quite elaborate. It comes
down to an analysis of the relation between S–polynomials and cancellation
of leading terms. The proof of ‘only if’ is a trivial consequence of the division
algorithm. A proof of the implication ‘if’ will be sketched below.
¤
1.6.9
A first indication of the usefulness of S–polynomials is given in the following
lemma.
P
1.6.10 Lemma. Let f = i ci X ai fi be a sum whose multidegree is less than d. If
ai + multdeg(fi ) = d for all i, then f can be written as a sum
f=
X
cj,k X d−cj,k S(fj , fk ),
j,k
where X cj,k is the least common multiple of the leading monomials of fj
and fk , and where the multidegree of each term in the sum is less than d.
Proof (sketch). We will demonstrate the proof in the case of two polynomials
f1 and f2 of multidegrees b1 and b2 , respectively. For simplicity we’ll also
assume that the leading terms of f1 , f2 have leading coefficient 1. Since the
multidegree of f is less than d, we have c1 + c2 = 0. So we can (re)write
f
= c1 (X a1 f1 − X a2 f2 )
d
= c1 ( XXb1 f1 −
Xd
f )
X b2 2
c
= c1 X d−c ( XXb1 f1 −
Xc
f )
X b2 2
= c1 X d−c S(f1 , f2 ),
as desired. In the third equality we use that X c divides X d ; this is the
case since Xia divides X d for i = 1, 2. For the general proof we refer to the
exercises.
¤
1.6.11 (Proof of Theorem 1.6.8) Again we restrict to the case that the basis
consists of two elements g1 , g2 . Among the (possibly many) ways to write
f = h1 g1 + h2 g2 , choose one in which the maximum of the multidegrees of
h1 g1 and h2 g2 is minimal, say d. Of course, multdegree(f ) ≤ d.
Now suppose first that the multidegree of f is strictly less than d. Then
the leading terms of h1 g1 and h2 g2 both have multidegree d and they cancel
1.6 Buchberger’s algorithm
11
each other. By Lemma 1.6.10, the sum lt(h1 )g1 + lt(h2 )g2 can be rewritten
using the S–polynomial S(g1 , g2 ):
lt(h1 )g1 + lt(h2 )g2 = cX d−c12 S(g1 , g2 ),
where c12 is the exponent of the least common multiple of g1 , g2 and where
the multidegree of the terms on the right–hand side is less than d. By
hypothesis, the S–polynomial S(g1 , g2 ) can be written as a sum a1 g1 + a2 g2
with multidegrees of the two terms bounded by the multidegree of S(g1 , g2 ).
Then use this to rewrite the original expression for f in such a way that
the degree d goes down. This contradicts our assumption. So, from f =
h1 g1 + h2 g2 , we deduce that the multidegree of f equals the multidegree of
at least one of the hi gi . But this implies that lt(f ) is divisible by the leading
term of the corresponding gi . This shows that lt(f ) ∈ (lt(g1 ), lt(g1 )).
¤
1.6.12 (Buchberger’s algorithm I) A first version of Buchberger’s algorithm is
easily described using the above S–polynomials. It is primitive in the sense
that no care is taken of efficiency matters.
We start with a basis (f1 , . . . , ft ) of the nonzero ideal I and transform this
basis stepwise into a Gröbner basis. In each step, we find an intermediate
finite basis G0 for the ideal, form all the possible S–polynomials of elements
in G0 . If division of such an S–polynomial by G0 leaves a nonzero remainder,
we add this remainder to our intermediate basis. If all these remainders are
zero, we stop and output the basis found so far; by Theorem 1.6.8 it is a
Gröbner basis. Of course, we need to show that the algorithm terminates
and then produces a Gröbner basis. Here is the algorithm schematically:
Input: F = (f1 , . . . , ft )
Output: a Gröbner basis g1 , . . . , gs
G=F
repeat
G0 = G
For each pair p, q in G0
do compute the S–polynomial S(p, q) and its remainder r(p, q)
upon division by G0
if r(p, q) 6= 0, then G := G ∪ {r(p, q)}
until G = G0
1.6.13 (Termination) To show that the algorithm terminates and upon termination produces a Gröbner basis, consider what happens if a nonzero remainder
12
Polynomials, Gröbner bases and Buchberger’s algorithm
R of an S–polynomial is added to G: G0 = G ∪ {R}. Then the leading term
of R is not divisible by any of the leading terms of the elements in G and
so the ideal (lt(G)) is strictly contained in the ideal (lt(G0 )) (here we use
Lemma 1.5.2 again). Since every ascending chain of ideals must eventually
become constant, the algorithm must terminate.
When the algorithm terminates in stage G, say, all remainders of S–
polynomials upon division by G are 0, and therefore G must be a Gröbner
basis by Theorem 1.6.8.
1.6.14
Many improvements can be made, but we refrain from doing this here. In
the computer algebra packages Mathematica and Maple, versions of Buchberger’s algorithm are available. From easy examples, it already becomes
claer that doing computations by hand is a very unpleasant task. So do try
to use computer algebra packages for such computations.
Chapter 2
Applications
2.1
Elimination
For systems of linear equations, the Gauss–elimination algorithm does exactly what the term suggests: in the process variables are eliminated from
the consecutive equations. A similar result holds for Gröbner bases if you
use the lex order.
2.1.1 Theorem. (Elimination) Let G be a Gröbner basis for the ideal I in
k[X1 , . . . , Xn ] with respect to lex order, where X1 > · · · > Xn . For j =
1, . . . , n, let Gj = G ∩ k[Xj+1 , . . . , Xn ]. Then Gj is a Gröbner basis for the
ideal Ij = I ∩ k[Xj+1 , . . . , Xn ].
Proof. If g 6∈ Gj , then the leading term of g must involve at least one of the
variables X1 , . . . , Xj : otherwise, the leading term would be in k[Xj+1 , . . . , Xn ],
and consequently, since we are using lex order with X1 > · · · > Xn , all
the other terms of g must be in k[Xj+1 , . . . , Xn ]. This would imply that
g ∈ k[Xj+1 , . . . , Xn ], a contradiction.
Suppose Gj = (g1 , . . . , gm ) ⊂ G = (g1 , . . . , gt ) and let f ∈ Ij . Then
division by G yields an expression f = q1 g1 +· · ·+qt gt . But since f, g1 , . . . , gm
do not involve the variables X1 , . . . , Xj , the quotients qm+1 , . . . , qt must all
be 0 (for instance, the leading term of gm+1 cannot divide the leading term
of f by what we remarked above).
So f ∈ (g1 , . . . , gm ) and division by G comes down to division by Gj .
Now every S–polynomial of two polynomials in Gj belongs to Ij (check!)
and the remainder upon division by Gj equals the remainder upon division
by G. The last mentioned remainder is 0 because G is a Gröbner basis (see
Theorem 1.6.8).
¤
13
14
Applications
2.1.2 Example. Suppose we have a curve in k 2 described parametrically by
x = t2 ,
y = t3 .
To find an equation for this curve, we want to eliminate t. In this example
this is quite easy to do ‘by hand’: x3 = y 2 . But in the process of computing
a Gröbner basis this equation occurs necessarily as a by–product. Fix the
lex order with t > x > y and consider the ideal I = (f1 = t2 −x, f2 = t3 −y).
Then the leading term of S(f1 , f2 ) = −(tx−y) is not divisible by the leading
terms of f1 and f2 , so we add f3 = tx − y to the generators of the ideal:
I = (f1 , f2 , f3 ). Next we compute S(f1 , f3 ) = ty − x2 . Again, the leading
term is not divisible by the leading terms of f1 , f2 , f3 . So we add f4 =
ty − x2 to the generators and turn to S(f1 , f4 ) = tx2 − xy. Since this equals
(tx − y)x, the remainder upon division by f1 , . . . , f4 is 0; the same holds for
S(f2 , f3 ) = t2 y − xy as we see from writing it as (t2 − x)y. Then we turn to
S(f2 , f4 ) = t2 x2 −y 2 . Upon division we get t2 x2 −y 2 = (t2 −x)x2 +(x3 −y 2 ),
leaving f5 = x3 − y 2 as remainder. Further S–polynomials yield no new
generators, so we obtain the following Gröbner basis for I:
f1 = t2 − x, f2 = t3 − y, f3 = tx − y, f4 = ty − x2 , f5 = x3 − y 2 .
The basis is not reduced, since the leading term of f2 is still divisible by
the leading term of f1 , so we may leave out f2 . (In fact, to speed up the
computation we should have done this as soon as we had f3 at our disposal.)
This leaves us with the reduced Gröbner basis
f1 = t2 − x, f3 = tx − y, f4 = ty − x2 , f5 = x3 − y 2 .
The elimination theorem states that I ∩ k[x, y] = (x3 − y 2 ). So every polynomial g(x, y) that becomes the zero–polynomial upon substituting x = t2
and y = t3 must be a multiple of x3 − y 2 .
There is a subtle problem here: from the above we see that the zeroset
V (x3 − y 2 ) contains the set of points of the form (t2 , t3 ), but it could be that
this last set is strictly contained in the first set. This is one of the reasons for
a more detailed analysis of sets of the form V (I). This is done extensively
in algebraic geometry.
2.1.3
Other applications of elimination are discussed in Section 2.4 and in Chapter ??.
2.2 Geometry theorem proving: first glimpse
2.2
15
Geometry theorem proving: first glimpse
2.2.1 The framework of ideals and Gröbner bases can be used to explore classical
geometry. The set–up works (in principle) when both the hypotheses of a
geometric situation and the statement about it can be expressed as polynomial equations. Before such equations can be deduced, a coordinate frame
has to be chosen, and the various objects have to be described in terms of
these coordinates.
2.2.2 Example. To illustrate this in a simple example, consider the following
situation. Draw a circle C and draw a line through the center of C. The
points of intersection of the line with C are denoted by P and Q. Choose a
third point R on the circumference of the circle. The claim is that P R and
QR are perpendicular.
R
P
Q
A circle with radius r > 0 is described by x2 + y 2 − r2 = 0. Two antipodal
points on the circumference are given by P = (−r, 0) and Q = (r, 0). The
third point on the circumference is R = (u, v). Since R is supposed to be
on the circle, this produces one hypothesis:
Hypothesis: u2 + v 2 − r2 = 0.
The statement to be proved is that (u+r, v) and (u−r, v) are perpendicular.
Using the standard innerproduct, this means:
Thesis: (u + r) · (u − r) + v · v = 0.
Now the thesis is easily seen to reduce to u2 + v 2 − r2 = 0, an equality that
holds because of the assumption we made.
In terms of ideals, we phrase this as follows. The hypothesis describes
the subset {(u, v, r) ∈ k 3 | u2 + v 2 − r2 = 0}, i.e., the vanishing locus V (I) of
the ideal I = (u2 + v 2 − r2 ). Now we want our thesis to hold on this subset,
i.e., we want the polynomial T = (u + r) · (u − r) + v · v to vanish on V (I).
16
Applications
A sufficient condition for T to vanish on V (I) is that it belongs to I. This
condition is trivially satisfied in our situation.
It is also trivial to conclude that an arbitrary point R = (u, v) such
that the two segments P R and QR are perpendicular lies on the circle with
radius r passing through P and Q.
2.2.3
In general, if the hypotheses are described by the polynomials h1 , . . . , hl
and the thesis is described by the polynomial T , then the thesis holds if
T ∈ (h1 , . . . , hl ). This way the problem is shifted to the ideal membership
problem.
Unfortunately, this condition is only sufficient and not necessary. For
example, if T 6∈ I but T 2 ∈ I, then T still vanishes on V (I). A precise
statement concerning this problem is given in the next section.
The following example shows another type of difficulty, which can arise
if one deals with geometry statements.
2.2.4 Example. Consider a right triangle ABC, whose angle at A is 90◦ . The
foot of the altitude from A on BC is called H. The statement is that H and
the midpoints P (of AB), Q (of BC), R (of AC) of the three sides lie on a
circle (this is called the circle theorem of Apollonius).
C
R
A
H
Q
P
B
Before we go into a proof using Gröbner bases, we sketch the classical
approach. The vertex A and the three midpoints P, Q, R form a rectangle
and lie on a circle (AQ and P R pass through the center of the circle) by the
previous example. Again by the previous example, since the triangle AHQ
has a right angle at H, the point H is on the circle with diameter AQ.
Now we turn to the description in terms of polynomials. We choose
coordinates in such a way that A is in the origin, that B = (2x, 0) and
C = (0, 2y). Then P = (x, 0), Q = (x, y) and R = (0, y). The circle passing
2.2 Geometry theorem proving: first glimpse
17
through P, Q, R has equation
1
1
1
(X − x)2 + (Y − y)2 − (x2 + y 2 ) = 0.
2
2
4
(For the sake of the example, we assume this as an ‘obvious’ step.) Now the
point H = (p, q) satisfies two conditions:
a) (p, q) ⊥ (−x, y), i.e., yq − px = 0;
b) (p, q) is on the line BC with equation yX + xY − 2xy = 0 yielding
yp + xq − 2xy = 0.
Therefore our hypotheses ideal becomes
Hypotheses: (yq − px, yp + xq − 2xy).
Next we compute a Gröbner basis with respect to lex order with x > y >
p > q to find
G := [ −y p − x q + 2 x y, −y q + p x, −y q 2 − y p2 + 2 y 2 q ].
We want to check whether p2 − xp + q 2 − yq is in the ideal. Doing a division
with remainder shows that this is not the case. But we do come close.
Consider
−q(−y q + p x) − p(−y p − x q + 2 x y) = p2 y + q 2 y − 2xyp = (p2 + q 2 − 2xp)y.
Adding −y(y q − p x) from the ideal yields (p2 + q 2 − yq − xp)y. This is the
one we need, except for a factor y. We would like to cancel out the factor
y, assuming that it is never 0. Algebraically, this can be done as follows:
extend the polynomial ring to k[x, y, p, q, t] and extend the ideal to
(yq − px, yp + xq − 2xy, 1 − yt) ⊂ k[x, y, p, q, t].
This time we have more luck: p2 + q 2 − yq − xp turns out to belong to this
ideal: a Gröbner basis (with respect to lex order, x > y > p > q > t) for the
ideal
I1 := (yq − px, yp + xq − 2xy, 1 − yt)
is
[−y p − x q + 2 x y, 2 p x − p2 − q 2 , t x q − 2 x + p, −q 2 + 2 y q − p2 ,
−1 + y t, −2 q + t q 2 + t p2 ],
18
Applications
and the remainder upon division turns out to be 0, confirming that p2 +
q 2 − yq − xp belongs to the new ideal. The interpretation is that we have to
assume that y 6= 0 to get a valid statement. This is not surprising once you
realize that in classical geometry the pictures drawn often implicitly assume
that certain quantities are nonzero.
Note also that f = f (1 − yt) + f yt with both terms on the right-hand
side in the extended ideal once f y belongs to I. This is the algebraic reason
why the above works.
2.2.5
The example shows that Gröbner bases can be of help in proving geometry
theorems, but also that things are not fully automatic. The reader has lots of
choice in assigning coordinates, but may run into various types of problems.
One of them is that the theorem one wants to prove tacitly requires the
situation to be ‘nondegenerate’.
The various projects with this course deal with some of these aspects.
2.3
2.3.1
The Nullenstellensatz
There is one aspect in the above considerations that we would like to elaborate on, since it is related to the famous Hilbert Nullstellensatz, a cornerstone
of modern algebraic geometry. It works only over algebraically closed fields,
like C. Usually the theorem is split in two parts. In the second statement
we come across the notion of the√radical of an ideal. Given an ideal I in a
N
ring R, the radical, denoted by I, is the ideal
√ {f ∈ R | f ∈ I for some
N > 0}. It is straightforward to check that I is indeed an ideal.
2.3.2 Theorem. (Weak Nullstellensatz) If I is an ideal in the polynomial ring
k[X1 , . . . , Xn ] over an algebraically closed field k such that V (I) = ∅, then
I = k[X1 , . . . , Xn ].
Proof. We refer to the exercises for the proof of this theorem.
¤
2.3.3 Theorem. (Hilbert’s Nullstellensatz) Let I be an ideal in k[X1 , . . . , Xn ],
where k is algebraically√closed. If f ∈ k[X1 , . . . , Xn ] vanishes identically on
the set V (I), then f ∈ I, i.e., there exists an m > 0 such that f m ∈ I.
Proof. Let I = (f1 , . . . , fm ) and assume for simplicity that f 6= 0. Consider
the ideal J = I + (1 − f Y ) ⊂ k[X1 , . . . , Xn , Y ]. If (a1 , . . . , an , an+1 ) ∈
k n+1 is in V (J), then f1 (a1 , . . . , an ) = · · · = fm (a1 , . . . , an ) = 0, so that
(a1 , . . . , an ) ∈ V (I) and 1−f (a1 , . . . , an )an+1 = 0, so that f (a1 , . . . , an ) 6= 0.
This contradicts the assumption that f vanishes on V (I). So V (J) = ∅.
2.4 Algebraic numbers
19
By the Weak Nullstellensatz J = k[X1 , . . . , Xn , Y ] and, consequently,
there exists a relation of the form
1 = g1 f1 + · · · + gm fm + g(1 − f Y )
for some g, g1 , . . . , gm ∈ k[X1 , . . . , Xn , Y ]. Now substitute 1/f for Y in the
above relation to find a relation of the form
1 = g1 (X1 , . . . , Xn , 1/f )f1 + · · · + gm (X1 , . . . , Xn , 1/f )f1 .
Multiplying through by a sufficiently high power of f , we find
f N = g̃1 f1 + · · · + g̃m fm
as claimed.
¤.
2.3.4 The weak Nullstellensatz generalizes the fact that over an algebraically
closed field every nonconstant polynomial in one variable has (at least) one
zero. Let I 6= (0) be an ideal in k[X]. Then I is generated by one element, say I = (f ). If I 6= k[X], i.e., if f is not a constant, then the weak
Nullstellensatz claims that f has a zero.
2.4
Algebraic numbers
2.4.1 In this section we discuss algebraic numbers and a way to compute their
minimal polynomials.
2.4.2 Definition. Let α ∈ C. Then α is called algebraic if there is a nonconstant
polynomial f (X) ∈ Q [X] such that f (α) = 0. The subset of algebraic
numbers is denoted by A. In lemma 2.4.11 we show that A is in fact a
subfield of C.
2.4.3 Example. Every rational number r is of√course algebraic: r is a zero of
X − r ∈ Q [X]. The nonrational number 2 is algebraic, since it is a zero
of X 2 − 2 ∈ Q [X].
√
The complex number α = 2 + i is also algebraic. To find√a polynomial
f such that f (α) = 0, we proceed as follows. From α − i = 2 we deduce
that (α − i)2 = 2. Working out this expression and rewriting a bit, we find
α2 − 3 = −2iα.
Squaring again yields
(α2 − 3)2 = −4α2 .
20
Applications
From this equality we conclude that α is a zero of X 4 − 2X 2 + 9 ∈ Q [X].
The numbers e and π are known to be non–algebraic.
2.4.4
Given α ∈ C \ {0}, the ideal Iα = {f ∈ C [X] | f (α) = 0} is generated by
one element p(X). If p(X) is not the zero polynomial, then p is necessarily
irreducible, for if p(X) = p1 (X) p2 (X), then from p1 (α) p2 (α) = 0 it follows
that either p1 or p2 (or both) belongs to Iα , so that one of the two factors is
constant. The unique polynomial with leading coefficient 1 that generates
Iα is called the minimal polynomial of α.
The case Iα = (0) corresponds to the situation that α is not an algebraic
number.
√
2.4.5 Example. The minimal polynomial of i + 2 is p(X) = X 4 − 2X 2 + 9. In
the previous example we noted that α is a zero of p. It remains to show
√ that
p is irreducible in Q [X]. The four roots of the polynomial are ±i ± 2 and
it is easy to see from these roots that any linear or quadratic factor of p is
not in Q [X].
2.4.6
Given an algebraic number α ∈ C, the field Q (α) consists of all expressions
of the form
g(α)
,
h(α)
where g, h are polynomials and where h(α) 6= 0. The relation with the ideal
Iα is as follows. Define the map F : Q [X] → Q(α) by q(X) 7→ q(α). Then
the kernel is precisely Iα and we get an induced injective map
F : Q [X]/Iα → Q (α).
Since Iα is generated by an irreducible polynomial, the ideal is maximal and
the quotient is a field. Therefore, the image is also a field contained in Q (α).
But this image contains α (the image of the class of X). As Q (α) is the
smallest field containing α, this implies that F is an isomorphism. It also
implies that Q (α) = Q [α]. The argument given here is a special case of the
fact that k[X]/(f (X)) is a field and can be identified with k(X), where X
is the class of X, if f (X) is irreducible and k is a field.
Now suppose Iα = (p(X)) where p(X) has degree n. As a Q–vectorspace,
Q [X]/Iα has the basis 1, X, X 2 , . . . , X n−1 . In terms of Q [α] this means that
1, α, . . . , αn−1 is a Q–basis of Q [α]. The integer n is called the degree of the
extension Q ⊂ Q(α) and is often denoted by [Q(α) : Q]. It is also common
usage to denote the extension as Q(α) : Q.
2.4 Algebraic numbers
21
In general, if L : K is an extension of fields, we denote by [L : K] the
dimension of L as K–vectorspace. Again, this dimension is called the degree
of the extension.
The discussion shows the following proposition:
2.4.7 Proposition. If α is algebraic with minimal polynomial of degree n, then
Q (α) = Q [α], and this field is a finite dimensional Q–vectorspace of dimension n, with Q–basis 1, α, α2 , . . . , αn−1 .
2.4.8 The equality Q [α] = Q (α) implies that every element of the field Q (α)
can be expressed as a polynomial in α. For instance, if α2 6= −1, there are
rationals a0 , . . . , an−1 such that
1/(α2 + 1) = a0 + a1 α + · · · + an−1 αn−1 ,
although we have not indicated a way to compute these coefficients.
If α and β are bot algebraic, then a similar equality,
Q (α, β) = Q [α, β],
holds. The proof of this statement is similar to the proof in the case of
one algebraic number and is based on considering the map Q (α)[X] →
Q (α)(β) = Q (α, β), determined by sending X to β.
In general, if α1 , . . . , αn are algebraic, then Q (α1 , . . . , αn ) = Q [α1 , . . . , αn ].
2.4.9 Theorem. Let L : K be subfields of C and suppose dimQ (K) = m and
dimK (L) = n. Then dimQ (L) = mn.
Proof. Suppose β1 , . . . , βn is a K–vectorspace basis of L and that α1 , . . . , αm
is a Q–vectorspace basis of K. We claim that the set of mn products αi βj
is a Q–vectorspace basis of L.
First we show that the set spans L over Q. If β ∈ L, then there exist
elements λ1 , . . . , λn ∈ K with
β = λ 1 β1 + · · · + λ n βn ,
since the βj ’s span L over K. Now every λi can be written as a Q–linear
combination of the α’s:
λi = ai1 α1 + · · · + aim αm
(i = 1, . . . , n).
Merging these expressions yields
XX
X
β=
(
aij αj )βi =
aij (αj βi ).
i
j
i,j
22
Applications
Next we turn to the linear independence of the elements. So suppose
X
µij αi βj = 0,
i,j
where alle the µij ∈ Q. Upon rewriting this as
XX
(
µij αi ) βj = 0
j
i
and using the independence of the βj over K we conclude
X
µij αi = 0 for all i.
i
Using the independence of the αi over Q, we finally conclude that µij = 0
for all i and j.
¤
2.4.10 Remark. We have stated the theorem in the context of subfields of C,
but it holds in general for field extensions K ⊂ L ⊂ M : if M is a L–
vectorspace of dimension m and L is a K–vectorspace of dimension n, then
M is a K–vectorspace of dimension mn. The proof given above, with minor
modifications, also works in this general context. The following result is
again a special version of a result valid in a more general context.
2.4.11 Lemma.
a) Let α, β be algebraic numbers. Then
[Q (α, β) : Q (α)] ≤ [Q (β) : Q ].
In particular, [Q (α, β) : Q ] ≤ [Q (α) : Q ] · [Q (β) : Q ].
b) Let K be a subfield of C with dimQ (K) < ∞ and let γ ∈ K. Then γ
is algebraic.
c) A is a subfield of C.
Proof. a) Suppose [Q (β) : Q ] = n. Let x ∈ Q [α, β] = Q (α, β). Then
x can be written as a polynomial in α and β. Rewrite this polynomial
as f1 (α) + f2 (α)β + · · · + fm (α)β m , where the fi are polynomials in one
variable. Since any β j (j ≥ n) can be rewritten as a Q–linear combination
of 1, β, . . . , β n−1 , we see that x can be written as
x = g1 (α) + g2 (α)β + · · · + gn−1 (α)β n−1 ,
2.4 Algebraic numbers
23
i.e., a Q (α)–linear combination of the n − 1 elements 1, β, . . . , β n−1 . The
dimension as Q (α)–vectorspace is therefore at most n.
The last statement in a) is a consequence of Theorem 2.4.9.
b) Suppose dimQ (K) = m. Then the m + 1 elements 1, γ, γ 2 , . . . , γ m must
be linearly dependent over Q. So there exist rational numbers a0 , . . . , am ,
not all of them zero, such that a0 + a1 γ + · · · + am γ m = 0. This expresses
that γ is an algebraic number.
c) Given algebraic numbers α and β, we must show that α − β and 1/α
(if α 6= 0) are also algebraic. Now all these elements belong to Q (α, β),
which is a finite extension of Q by a). Then b) implies that every element
in Q (α, β) is algebraic.
¤
2.4.12 It is common practice to draw field extensions in pictures like the following.
√ √
Q ( 3 2, 3)
√
Q ( 3)
√
Q ( 3 2)
2
3
Q
The edges are often labelled with the degrees (known so far) of the corresponding extensions.
√
√
3
2,
3) : Q ] use the lemma√to √
de2.4.13 Example. To√compute
the
degree
[Q
(
√
√
duce that [Q ( 3 2, 3) : Q ( 3 2)] ≤ 2. Theorem 2.4.9 implies that [Q ( 3 2, 3) :
Q ] is divisible by 2 and 3 and that the total degree is bounded by 6:
√
√
√
√
√
√
3
3
3
3
[Q ( 2, 3) : Q ] = [Q ( 2, 3) : Q ( 2)] · [Q ( 2) : Q ] ≤ 2 · 3 = 6.
√ √
√ √
In conclusion, [Q ( 3 2, 3) : Q ] = 6. This implies, for example, that 3 2+ 3
has minimal polynomial of degree at most 6 (in fact, it turns out to have
degree exactly 6, see Example 2.4.16).
2.4.14 Next we turn to the computation of minimal polynomials of polynomial
expressions of two given algebraic numbers. Suppose α has minimal polynomial p(X) and β has minimal polynomial q(X). Suppose that we want
to compute the minimal polynomial of some polynomial expression f (α, β)
24
Applications
of α and β. Note that f (α, β) ∈ Q (α, β). (Lemma 2.4.11a) implies that
[Q (α, β) : Q ] < ∞ and b) of the same lemma implies that f (α, β) is algebraic.)
The abstract way to compute with α is to compute modulo p(X). Similarly, computing with β comes down to computing modulo q(X). To separate these two, we do the computations in Q [X, Y ]/(p(X), q(Y )). To compute with f (α, β) we introduce a third variable Z and add the generator
Z − f (X, Y ) to the ideal:
Q [X, Y, Z]/(p(X), q(Y ), Z − f (X, Y )).
So the class of Z represents our algebraic number f (α, β). To find its
minimal polynomial we compute a reduced Gröbner basis G for the ideal
I = (p(X), q(Y ), Z − f (X, Y )) with respect to lex order and such that Z
is the smallest variable. Then we consider G ∩ Q [Z]. By the elimination
theorem 2.1.1 this produces a generator for I ∩ Q [Z]. This generator is the
minimal polynomial of f (α, β). The formal statement (there is one condition
in the theorem that we neglected so far) and its proof are as follows.
2.4.15 Theorem. Let α and β be algebraic numbers with minimal polynomials
p(X), q(X), respectively. Suppose furthermore that q(X) is irreducible over
Q (α). Let f (X, Y ) ∈ Q [X, Y ]. Then the minimal polynomial of f (α, β) is
the generator of the ideal ((p(X), q(Y ), Z − f (X, Y )) ∩ Q [Z] in Q [Z].
Proof. We split the proof in several parts. As an auxiliary part, we first
establish that
Q [X, Y ]/(p(X), q(Y )) ∼
= Q(α, β),
under the natural map determined by X 7→ α, Y 7→ β. Start with the
isomorphism Q [X]/(p(X)) → Q (α) sending the class of X to α and use it to
define Φ : Q (α)[Y ] → Q (α, β), determined by φ2 (Y ) = β. This morphism is
clearly surjective. The kernel consists precisely of those polynomials h(Y ) ∈
Q (α)[Y ] such that h(β) = 0. We were given that q(Y ) ∈ ker(Φ). Now the
ideal ker(Φ) is generated by a single element. From the irreducibility of q(Y )
we deduce that ker(Φ) = (q(Y )). So
Q (α)[Y ]/(q(Y )) ∼
= Q (α, β).
Now use Q (α) ∼
= Q [X]/(p(X)) to conclude that Q [X, Y ]/(p(X), q(Y )) ∼
=
Q (α, β) (further details in the exercises).
Having established this, we turn to the actual proof. Consider the ring
morphism determined by
φ : Q [Z] → Q [X, Y ]/(p(X), q(Y )),
Z 7→ f (X, Y ) + (p(X), q(Y )).
2.4 Algebraic numbers
25
Then g(Z) is in the kernel of this map if and only if f (α, β) = 0. We need
to show that this kernel equals ((p(X), q(Y ), Z − f (X, Y )) ∩ Q [Z]. Consider
the following maps
j
ψ
Q [Z] ,→ Q [X, Y, Z] −→ Q [X, Y ]/(p(X), q(Y )) ∼
= Q (α, β),
where ψ maps X and Y to their respective classes and maps Z to the
class of f (X, Y ). The kernel of the map ψ evidently contains p(X), q(Y ),
Z − f (X, Y ). It is in fact precisely I. To see this, start with F (X, Y, Z) ∈
ker(ψ), and rewrite F as a polynomial in Z with coefficients in Q [X, Y ]:
a0 (X, Y ) + a1 (X, Y )Z + · · · + am (X, Y )Z m .
Now replace every occurrence of Z by (Z − f (X, Y )) + f (X, Y ) and expand
accordingly (formally, replace Z by U + V and expand; then substitute
Z − f (X, Y ) for U and f (X, Y ) for V and do not expand any further). The
result is that F is rewritten as the sum of a multiple of Z − f (X, Y ) and
and a polynomial in X, Y :
(Z − f (X, Y ))h(X, Y, Z)+
a0 (X, Y ) + a1 (X, Y )f (X, Y ) + · · · + am (X, Y )f (X, Y )m .
Since F and (Z − f (X, Y ))h(X, Y, Z) are in the kernel of ψ, we conclude
that a0 (X, Y ) + a1 (X, Y )f (X, Y ) + · · · + am (X, Y )f (X, Y )m is also in the
kernel of ψ. But this polynomial maps to the class of
a0 (X, Y ) + a1 (X, Y )f (X, Y ) + · · · + am (X, Y )f (X, Y )m
in Q [X, Y ]/(p(X), q(Y )). So
a0 (X, Y ) + a1 (X, Y )f (X, Y ) + · · · + am (X, Y )f (X, Y )m ∈ (p(X), q(Y )).
Altogether we find that
F (X, Y, Z) = (Z − f (X, Y ))h(X, Y, Z) + a0 (X, Y )+
a1 (X, Y )f (X, Y ) + · · · + am (X, Y )f (X, Y )m
∈ ((p(X), q(Y ), Z − f (X, Y )).
Now the kernel of φ is the kernel of the composition ψ◦j, where j denotes the
inclusion map. This means that g(Z) ∈ ker(φ) if and only if g(Z) ∈ ker(ψ),
i.e., g(Z) ∈ I (note that we identify Q [Z] with its image under j).
¤
26
Applications
√
2.4.16 Example.√The minimal polynomial of 2 is X 2 − 2 and the minimal
√
√polynomial of 3 2 is X 3 − 2. To find the minimal polynomial of 2 + 3 2 we
compute a Gröbner basis for the ideal (X 2 − 2, Y 3 − 2, Z − X − Y ) in
Q [X, Y, Z] with respect to the lex order, where X > Y > Z. The result is
[310 X − 24 Z 5 − 462 Z + 156 Z 2 − 9 Z 4 + 364 + 160 Z 3 ,
152 Z − 156 Z 2 + 9 Z 4 + 310 Y − 364 − 160 Z 3 + 24 Z 5 ,
−24 Z + 12 Z 2 − 6 Z 4 − 4 − 4 Z 3 + Z 6 ].
The last polynomial is one
only Z and by the above must be the
√ involving
√
minimal polynomial of 2 + 3 2:
−24 Z + 12 Z 2 − 6 Z 4 − 4 − 4 Z 3 + Z 6 .
2.4.17 Remark. There are many variants on this theorem for computing minimal
polynomials of various combinations of algebraic numbers.
Chapter 3
Factorisation of polynomials
3.1
Introduction
This chapter deals with the problem of factoring polynomials into irreducible
factors. We will sketch approaches to the the problem of factoring polynomials in Fq [X], where q is a prime power, and of polynomials with integer
coefficients. We begin with a section on generalities.
3.2
Polynomials with integer coefficients
3.2.1 If k is a field, then every polynomial in k[X] factors into irreducibles in
a (up to the order of the factors and up to constant factors) unique way.
This holds in particular for polynomials in Q [X]. In the following it will
be essential to work with polynomials with integer coefficients. Fortunately,
factors of a polynomial with integer coefficients can always be taken in Z [X].
To explain this, we denote by c(f ) the gcd of the coefficients of f ∈ Z [X],
the so–called content of f . The first statement in the following proposition
is usually called the Lemma of Gauss.
3.2.2 Proposition.
a) The content is multiplicative in the sense that for nonzero polynomials f, g ∈ Z [X] the relation c(f g) = c(f ) c(g) holds.
b) If f ∈ Z [X] factors as f = gh with g, h ∈ Q [X], then there exist
a, b ∈ Q such that ag, bh ∈ Z [X] and f = (ag) (bh).
Proof. a) The crucial case to consider is when c(f ) = c(g) = 1. Reducing
modulo any prime p we get f g = f · g with f , g 6= 0. As Fp [X] is an integral
27
28
Factorisation of polynomials
domain, we conclude that f g 6= 0 showing that c(f g) has no prime divisors,
hence equals 1.
b) The proof is easily reduced to the case where c(f ) = 1. In that
case multiply g and h by rational numbers a and b, respectively, in such a
way that ag and bh have integer coefficients and content 1. Then c(abf ) =
c(ag)c(bh) = 1 by a), so that ab = ±1. Altering the sign of a if necessary,
we can arrange it so that ab = 1 and f = (ag) (bh).
¤
3.2.3
Now an obvious operation on polynomials with integer coefficients is to
reduce their coefficients modulo a prime. Let p be a prime and let f =
am X m +am−1 X m−1 +· · ·+a1 X +a0 be a polynomial with integer coefficients.
Then the reduction mod p of f is the polynomial
f = am X m + am−1 X m−1 + · · · + a1 X + a0 .
3.2.4 Proposition. Let f ∈ Z [X] of positive degree whose leading coefficient is
not divisible by the prime p. If the reduction of f mod p is irreducible, then
f is irreducible as a polynomial in Q [X].
Proof. Suppose f factors as f = gh with g and h of positive degree. By
the Lemma of Gauss, we may assume that g and h have integer coefficients.
Upon reducing mod p we find the equality f = g · h in Fp [X], contradicting
the irreducibility of f .
¤
3.2.5
The next criterion is equally simple to prove, yet is much subtler.
3.2.6 Proposition. (Eisenstein’s criterion) Let p be a prime and let f =
am X m +am−1 X m−1 +· · ·+a1 X +a0 be a polynomial with integer coefficients
satisfying:
a) p does not divide am ;
b) p divides am−1 , am−2 , . . . , a0 ;
c) p2 does not divide a0 .
Then f is irreducible in Q [X].
Proof. If f factors as f = gh with g and h both in Z [X] and of positive
degree, then reduction mod p yields f = g · h. Since f = am X m 6= 0 and
since we have unique factorization in Fp [X], we conclude that g and h each
consists of its leading term only. In particular, their constant terms are 0,
so that the constant terms of g and h are divisible by p. But then a0 is
divisible by p2 , a contradiction.
¤
3.3 Factoring polynomials modulo a prime
29
3.2.7 We leave the following as an exercise for the reader. Let k be a field and let
a ∈ k. If f ∈ k[X], then f is irreducible if and only if f (X +a) is irreducible.
3.2.8 Example. For p prime, define Φp (X) = X p−1 + X p−2 + · · · + X + 1. Then
Φp (X) =
Replace X by X + 1 and we find
Φp (X + 1) =
Xp − 1
.
X −1
(X + 1)p − 1
.
X
The right–hand side works out as
¶
µ ¶
µ
p
p
X + p.
X p−2 + · · · +
X p−1 +
2
p−1
This polynomial is suitable for the application of Eisenstein’s criterion for
the prime p. We conclude that Φp (X) is irreducible.
Φp (X) is part of a family of polynomials, the cyclotomic polynomials
Φm (X) for m ∈ Z, m > 0. These are the minimal polynomials of
e
2πi
m
and of fundamental importance in number theory.
3.3
Factoring polynomials modulo a prime
3.3.1 Factoring a polynomial in Fq [X] (with q a power of the prime p) is a finite
job. The purpose of this section is to demonstrate Berlekamp’s algorithm,
a more efficient way of factoring. It is based on the following observation.
3.3.2 Lemma. Let f ∈ Fq [X] and suppose u ∈ Fq [X] satisfies uq ≡ u (mod f ).
Then
f = Πa∈Fq gcd(f, u − a).
Proof. Substituting u in the equality X q − X = Πa∈Fq (X − a), we find
uq − u = Πa∈Fq (u − a). Since (u − a) − (u − b) = b − a is a constant, The
factors on the right–hand side are relatively prime. As f divides uq − u, we
find the following string of equalities
f = gcd(f, uq − u) = gcd(f, Πa∈Fq (u − a)) = Πa∈Fq gcd(f, u − a).
¤
30
Factorisation of polynomials
3.3.3 Example. Let f = X 4 − 1 ∈ F5 [X] and let u = X. Then the conditions of
the lemma are satisfied and we find the factors
gcd(f, X − 0) = 1,
gcd(f, X − 1) = X − 1, gcd(f, X + 1) = X + 1,
gcd(f, X − 2) = X − 2, gcd(f, X + 2) = X + 2.
In this example we happen to find the full factorisation. This need not be
the case in general.
3.3.4
Of course, when one u doesn’t work, another one may. So it makes sense
to vary the choice of u. To do this, we first investigate the structure of the
set of all u satisfying uq ≡ u (mod f ). It turns out to be an m–dimensional
vector space over Fq , where m is the number of distinct irreducible divisors
of f . Before we state this and give the proof, we first explain an ingredient
em . Consider the map
of the proof. Suppose f factors as f = f1e1 · · · fm
em )
φ : Fq [X] → Fq [X]/(f1e1 ) × · · · × Fq [X]/(fm
e1
e
g 7→ (g + (f1 ), . . . , gm + (fmm ))
It is easy to verify that this map is a morphism of rings. The kernel consists
e
of the polynomials h ∈ Fq [X] such that fj j divides h for j = 1, . . . , m. Since
e
the fj j are relatively prime, we conclude that h is in the kernel if and only
if f | h. But this implies that we get a well-defined injective morphism
em ),
φ : Fq [X]/(f ) → Fq [X]/(f1e1 ) × · · · × Fq [X]/(fm
e1
e
m
g + (f ) 7→ (g + (f1 ), . . . , gm + (fm )).
The number of elements in Fq [X]/(f ) is q deg(f ) and the number of elements
em ) is q e1 deg(f1 ) · · · q em deg(fm ) . Since these
in Fq [X]/(f1e1 ) × · · · × Fq [X]/(fm
numbers are equal the map φ is an isomorphism.
This isomorphism (a polynomial version of the Chinese remainder theorem) enables us to translate statements about Fq [X]/(f ) into statements
em ) and conversely. Note that it is not
about Fq [X]/(f1e1 ) × · · · × Fq [X]/(fm
evident how to describe the inverse of φ.
The class of an element u ∈ Fq [X] such that f | uq − u behaves as follows
under the isomorphism. The decomposition
uq − u = Πa∈Fq (u − a)
into relatively prime factors shows that each fj divides exactly one of these
e
factors, say u − a. But then fj j divides u − a because f |uq − u. So modulo
e
fj j , the class of u corresponds to a constant a. So the image of an element
3.3 Factoring polynomials modulo a prime
31
u satisfying f | uq − u belongs to the subring (Fq )m of Fq [X]/(f1e1 ) × · · · ×
em ).
Fq [X]/(fm
Conversely, since every element a ∈ Fq satisfies aq = a, every element of
the subring (Fq )m comes from an element u ∈ Fq [X]/(f ) satisfying uq = u.
This proves the following proposition.
3.3.5 Proposition. The set Sf = {u ∈ Fq [X]/(f ) | uq = u} is an m–dimensional
Fq –vector space, where m is the number of distinct irreducible divisors of f .
Under the isomorphism
em
φ : Fq [X]/(f ) → Fq [X]/(f1e1 ) × · · · × Fq [X]/(fm
)
the set Sf corresponds to the subring (Fq )m of constants.
3.3.6 To find the set Sf we need to solve a system of equations in Fq [X]/(f ).
Since u(X)q = u(X q ), solving uq = u (mod f ) comes down to solving a
system of linear equations.
e
Berlekamp’s algorithm to find the factors fj j of f runs as follows: first
determine a vector space basis u1 = 1, . . . , um of Sf and let D = {f }. In
each stage of the algorithm the set D contains polynomials whose product
equals f . If m = 1, there is only one prime power in f and we are done. If
m > 1, then we use Lemma 3.3.2 and replace for j = 2, . . . , m every element
g of D by the nontrivial elements among the gcd(g, uj − a) with a ∈ Fq .
Since only finitely many steps are involved, the algorithm terminates and
em }. We do this in two
it remains to check that at the end D = {f1e1 , . . . , fm
steps.
a) First note that in every step of the algorithm every element of D is
either divisible by fiei or is relatively prime to fi (for every i): if this
property holds for g, then it is inherited by every gcd(g, u − a), where
a ∈ Fq , since every u − a is either relatively prime to fi or is divisible
by fiei (see 3.3.4).
b) If g does not decompose further at the end, then for every j there exists
an sj ∈ Fq such that g | uj − sj . Since the uj make up a Fq –basis of
Sf , a similar statement holds for arbitrary u ∈ Sf : for every u ∈ Sf
there exists an su ∈ Fq such that g | u − su . If g is divisible by the
factors fkek and flel then the k–th and l–th components of φ(u) equal
su , contradicting the surjectivity of φ.
3.3.7 The algorithm so far produces the factors fiei of f . The last step consists of
finding the fi from these powers. Before we proceed, we recall the following
32
Factorisation of polynomials
equality for polynomials in Fq [X]:
g(X)p = g(X p )
k
k
and consequently g(X)p = g(X p ).
For example, in characteristic 2, we have
X 8 + X 4 + 1 = (X 2 )4 + X 4 + 14 = (X 2 + X + 1)4 .
Now let’s suppose we are given g n and we wish to find g. The first approach
is to differentiate g n , this yields (g n )0 = ng 0 g n−1 , and divide g n by this
derivative. (We remark that we haven’t defined differentiation in characteristic p and the following discussion shows exactly some of the pitfalls.) This
goes well if p does not divide n and if g 0 6= 0. If g 0 = 0, then all exponents in
g are divisible by p. By using the above equality repeatedly, we can absorb
all these factors p and rewrite g n as
k
g(X)n = h(X p )t ,
with p relatively prime with t and with at least one of the exponents of
h(X). Then compute h/h0 .
3.4
3.4.1
Factoring polynomials over the integers
As we saw above, factoring over the rationals is essentially the same as
factoring over the integers, so we will focus on the latter. We started the
discussion on factoring polynomials mod q by remarking that it is a finite
problem anyway. It takes a little consideration to show that factoring over
the integers or rationals is also a finite job: if g ∈ Z [X] divides f ∈ Z [X],
then it is easy to see that the coefficients of g can be bounded in terms
of those of f . Just start with the leading coefficient and work through the
coefficients one by one, bookkeeping the bounds found so far. Without proof
we state below a nice bound (in fact sharp), the Landau–Mignotte
bound.
P
To state it, we need the so–called l2 –norm of a polynomial f = ni=0 fi X i ∈
Z [X]:
v
u n
uX
||f || = t
fi2 .
i=0
Of course, our algorithmic approach will be better than just an exhaustive
search using this bound. In this section we restrict ourselves to a discussion
of an algorithm based on the Hensel lift.
3.4 Factoring polynomials over the integers
33
3.4.2 Theorem. (Landau–Mignotte) If f, g ∈ Z [X] are of degrees n and m,
respectively, and if g | f , then
µ ¶
m
||f ||
|gi | ≤
i
for i = 0, 1, . . . , m. A bound for the l2 –norm of g is given by
µ ¶2
2m
||f ||2 .
||g|| ≤
m
2
3.4.3 The starting point for the Hensel lift is a factorization of the reduction of a
polynomial modulo a prime p. The Hensel lift aims at lifting a factorisation
modulo pk to a factorisation modulo pk+1 . Together with the Landau–
Mignotte bound on the coefficients of possible factors, these ingredients can
be put together into an algorithm.
3.4.4 Theorem. (Hensel lift) Let p be a prime and let f, g, h ∈ Z [X] be monic
polynomials of positive degree such that f ≡ g h (mod pm ) for some m ∈ N
with gcd(g (mod p), h (mod p)) = 1. Then there exist monic polynomials
g̃, h̃ such that g̃ ≡ g (mod pm ), h̃ ≡ h (mod pm ) and f ≡ g̃ h̃ (mod pm+1 ).
Moreover, the lifts g̃ and h̃ are uniquely determined modulo pm+1 .
Proof. The polynomials g̃ and h̃ are required to be of the form
g̃ = g + pm u,
h̃ = h + pm v,
with the polynomials u and v satisfying deg(u) < deg(g) and deg(v) <
deg(h). Using these two explicit forms, the equation f ≡ g̃ h̃ (mod pm ) can
now be rewritten as
f − gh
− (uh + vg) ≡ 0 (mod p).
pm
Since gcd(g (mod p), h (mod p)) = 1, there exist polynomials u of degree
less than deg(g) and v of degree less than deg(h) such that
f − gh
≡ uh + vg
pm
(mod p).
Moreover, modulo p these polynomials u and v are unique. Consequently,
the resulting g̃ and h̃ are unique modulo pm+1 .
¤
34
Factorisation of polynomials
3.4.5 (Factorization) Let f ∈ Z [X] be square–free of degree n > 0 and suppose
c(f ) = 1. Choose un upper bound C for the coefficients of possible divisors
of f , for instance the Landau–Mignotte bound. Also choose a prime p that
does not divide the discriminant of f (so that the reduction of f mod p
is also square–free) and does not divide the leading coefficient an of f (so
that the reduction mod p does not drop in degree) and an integer m so that
pm > 2 |an | C. The algorithm then goes through the following steps:
a) Factor f mod p.
b) Lift this factorization to a factorization
f ≡ a n g1 · · · g k
(mod pk ),
with monic gi .
c) For every subset S of {1, 2, . . . , k} compute a polynomial h ∈ Z [X]
such that
(i) h ≡ an Πi∈S gi (mod pm );
(ii) h has degree at most [n/2];
(iii) the absolute values of the coefficients of h are less than |an |C.
Finally, test whether h divides f and assemble the divisors of f .
The number of tests in the final step can be exponential in n, for instance
if f factors into linear factors modulo p.
Chapter 4
Symbolic integration
4.1
Introduction
4.1.1 In this chapter we are concerned with the problem of finding exact antiderivatives or indefinite integrals: given a function f (x), find F (x) such
that
Z
F 0 (x) = f (x)
or
f (x) dx = F (x).
In basic calculus one learns a range of methods for determining indefinite
integrals, most notably integration by parts and the substitution rule. But
in most cases a fully algorithmic approach is not clearly available. One is
often guided by intuition and experience. Some integrals don’t seem to yield
to any method; for instance,
Z
2
e−x dx
is such an integral. In such cases, sometimes there do exist methods to
determine specific definite integrals, like
Z ∞
2
e−x dx,
0
but that will not be the topic of this chapter. We will be solely interested
in the problem of finding antiderivatives in the following two senses:
• given an expression f (x) in terms of ‘elementary’ functions, determine
(algorithmically) an antiderivative in terms of ‘elementary functions’,
or
35
36
Symbolic integration
• show that there is no antiderivative of f (x) in such elementary terms.
The term ‘elementary function’ will be made more precise below, but for
the moment it suffices to know that it refers to combinations of the usual
functions, like polynomials, sine, cosine, exponential, where combination is
meant in the sense of composition of functions and in the sense of applying
the usual arithmetical operations like addition, multiplication, taking n–th
roots. An example of an elementary expression is
p
5
ex2 + sin(x)
√ .
log(3 + x2 − sin( 4 ex )
4.1.2 Remark. In this chapter we shall be interested in formal properties of differentiation and integration; domains of definition of a given expression are
of no importance for us, i.e., we will be studying expressions rather than
functions, although our terminology will be sloppy in this respect and the
terms function and expression are both used.
4.1.3
From the following section on, all fields in this chapter have characteristic
zero.
4.2
4.2.1
Differential fields
From the point of view of differentiation, polynomials and rational functions are quite simple. In the algebraic context, differentiation can be viewed
as a certain operator on the field K(x) of rational functions over the field
K. In this algebraic setting more complicated expressions, like the logarithm, can be treated by adjoining them to the field K(x). For this to make
sense we need to be able to algebraically characterize such expressions and
to define how differentiation extends to this larger field.
The idea of using fields is somewhat reminiscent of Galois theory and
there is a striking resemblance between two of the corner stones in both
theories: solving polynomial equations by radicals in Galois theory versus
solving indefinite integrals in symbolic integration. We note that there is an
obvious extension of symbolic integration to differential equations, which is
often called differential Galois theory.
In the sequel we will work in suitable field extensions. Here, suitable
means that the field is relevant to our specific class of functions, but is also
adapted to the process of differentiation. The formal notion is the following.
4.2 Differential fields
37
4.2.2 Definition. A differential field consists of a field K of characteristic 0 and
a map D : K → K satisfying the rules
a) D(f + g) = Df + Dg for all f, g ∈ K;
b) (Leibniz’ rule) D(f · g) = f Dg + g Df for all f, g ∈ K.
The map D is a so–called derivation or differential operator. If D is understood, then we simply say ‘the differential field K’. If we need to be more
precise we write for example (K, D). Again, if no confusion arises, we also
write f 0 instead of D(f ).
4.2.3 Example. The standard example is the field of rational functions in one
variable over a field, say, the complex numbers or the rational numbers,
K = C (x)
or Q (x),
with derivative D determined by:
D(an xn + · · · + a2 x2 + a1 x + a0 ) = nan xn−1 + · · · + 2a2 x + a1 ,
D( fg ) =
g Df −f Dg
.
g2
The coefficients ai are constants in C or Q, respectively, the elements f, g
are polynomials over C or Q, respectively, with g 6= 0.
4.2.4 For the sake of practicality (think of implementations in computer algebra
systems), we usually focus on extensions of the field Q. There is a price to
pay for this starting point, namely that we have to accept that sometimes
algebraic integers have to be adjoined.
The proof of the following properties is left as an exercise.
4.2.5 Proposition. In the differential field (K, D), the following properties hold:
a) D(0) = D(1) = 0, more generally, D(r) = 0 for all r ∈ Q ⊂ K.
b) The set {c ∈ K | D(c) = 0} is a subfield of K. This field is called the
field of constants with respect to D.
c) D(af +bg) = aD(f )+bD(g) for all f, g ∈ K and all a, b ∈ K satisfying
D(a) = D(b) = 0.
d) D( fg ) =
g D(f )−f D(g)
g2
for all f, g ∈ K with g 6= 0.
38
Symbolic integration
e) D(f n ) = n f n−1 D(f ) for all f (6= 0) ∈ K and n ∈ Z.
4.2.6
Similar to field extensions in ordinary field theory are the differential field
extensions in our context. The important aspect is of course the role of the
differential operator.
4.2.7 Definition. Let (L, DL ) and (K, DK ) be differential fields, where L is a
field extension of K. If DL (f ) = DK (f ) for all f ∈ K, then (L, DL ) is called
a differential extension field of (K, DK ). If no confusion arises, we simply
say that L is a differential extension field of K.
4.2.8
For our purposes three types of extensions will play a crucial role; one
type of extension is of the sort discussed in the previous chapter; the two
other types correspond to adjoining a ‘logarithm’ or ‘exponential’, respectively. The definitions are inspired by the formal differentiation properties
of logarithms and exponentials. Again, we note that in the sequel it will be
irrelevant for us to view expressions as functions; domains of definition play
no role whatsoever.
4.2.9 Definition. (Logarithm, exponential) A differential field extension L :
K is said to contain a logarithm of u ∈ K with u 6= 0 if L contains an
element θ such that
D(u)
D(θ) =
.
u
We often write θ = log(u).
The extension L : K is said to contain an exponential of u ∈ if it contains
an element θ such that
D(θ) = D(u) · θ.
We often write eu or exp(u) for such θ.
4.2.10
Note that we haven’t proved that given u ∈ K an extension containing
a logarithm or exponential exists. Also, we are not claiming that such
logarithms or exponentials necessarily bring us outside the field K.
The notation log(u) and eu suggests that such expressions share more
of the properties of the logarithm and exponential, respectively. Indeed,
as regards the arithmetic structure of the field, we expect properties like
log(u) + log(v) = log(uv) and eu · ev = eu+v .
In the sequel we will be a bit sloppy about the precise construction of logarithmic and exponential extensions, and simply use the various properties
whenever convenient.
4.3 Rational functions
39
4.2.11 Before we turn to a specific class of functions in the next section, we end
this section by showing that logarithms occur more often in the process of
integration than one is inclined to think at first glance. Here, it will be an
advantage that we need not be concerned about domains of definition of our
expressions.
The integral
Z
1
= log(x)
x
shows that we need the logarithm, an example of a non–rational function
(see 4.3.2), to integrate a relatively easy expression like 1/x. In the integral
Z
1
= arctan(x)
2
x +1
the right–hand side suggests that we need again a different type of non–
rational function to express the antiderivative of 1/(x2 + 1). From the analyst’s point of view, the arctan may be the best way of writing the integral,
but for an algebraist the arctan obscures the structure of the integral in the
following sense. The identity
x2
i
1
1
1
=− (
−
)
+1
2 x+i x−i
enables us to rewrite the integral with the help of logarithms as
Z
i
1
dx = − (log(x + i) − log(x − i)),
2
x +1
2
at the expense of extending the coefficients to Q (i). In the exercises you
will show that more integrals, which are usually not expressed in terms of
logarithms, can be expressed in terms of logarithms.
In fact, for rational functions we will see in the next section that logarithmic extensions suffice in the integration process.
4.3
Rational functions
4.3.1 A class of functions where the collection of classical integration rules is
most successfull is the class of rational functions. In fact, partial fraction
expansion and the standard rules of integration applied to rational functions
come close to an algorithmic description. The purpose of this section is
to explain this algorithmic approach. We first show that integrating 1/x
necessarily brings us outside the realm of rational functions.
40
Symbolic integration
4.3.2 Theorem. Let Q (x) be the function field in one variable with the usual
differentiation D.
a) The field of constants in Q (x) is Q.
b) There exists no f ∈ Q (x) such that D(f ) = 1/x.
Proof. We leave the proof of a) as an exercise and turn to the proof of
b). If p and q are polynomials such that D(p/q) = 1/x, then the rules for
differentiation imply
x q D(p) − x p D(q) = q 2 ,
implying x|q 2 and therefore x|q. Upon rewriting q as xn q1 with x and q1
relatively prime and substituting we find
xn+1 q1 D(p) − nxn pq1 − xn+1 p D(q1 ) = x2n q12 .
From this equality we obtain x|p contradicting the assumption that p and q
are relatively prime.
¤
4.3.3
In integrating a rational function, the first step is usually to apply partial
fraction expansion. This reduces the problem to a set of usually simpler
problems. Below we describe the various types of fractions we would like to
distinguish and how to handle them. Following this discussion we show how
to use partial fraction expansion to put everything together.
The next lemma is used below; its proof is left as an exercise.
4.3.4 Lemma. (Integration by parts) For every u, v in a differential field, the
following equality holds
Z
Z
u D(v) = uv − v D(u).
4.3.5 (Square–free denominator) If the denominator b of the rational function a/b ∈ Q (x) (with a and b relatively prime, b monic, and deg(a) <
deg(b)) is squarefree, and if b factors as
b = (x − c1 )(x − c2 ) · · · (x − cn )
(the ci are distinct elements in C), then a/b can be written as a sum
n
a X αi
=
,
b
x − ci
i=1
4.3 Rational functions
41
with unique constants αi . Therefore, we have
Z
n
a X
αi log(x − ci ).
=
b
i=1
To be able to write down this expression, we need to adjoin the roots
c1 , . . . , cn to the field Q and adjoin the various log(x − ci ). Since the roots
are algebraic, adjoining them involves at most a finite extension of Q. If two
or more of the coefficients αi coincide, then we can rearrange the sum of the
logarithms. For instance, if α1 = α2 , then
α1 log(x − c1 ) + α2 log(x − c2 ) = α1 log((x − c1 )(x − c2 )).
This rearranging may reduce the degree of the algebraic extension we need
to write down the integral as the following example shows.
Z
Z
2x
1
1
=
+
= log(x + i) + log(x − i) = log(x2 + 1).
x2 + 1
x+i x−i
In this case no constants outside the rationals are necessary. But we still
need a logarithm.
4.3.6 (Denominator is a pure power) Here we are dealing with the case a/bm
with
• deg(a) < deg(b),
• b is squarefree,
• m > 1.
The idea in this case is to apply integration by parts to reduce the exponent
in a/bm as far as possible, in fact to 1. Since b is squarefree, gcd(b, D(b)) = 1,
and the Euclidean algorithm produces an identity of the form
ub + vD(b) = a.
Dividing both sides of the equation by bm and integrating we get
Z
Z
Z
a
u
v D(b)
=
−
.
bm
bm−1
bm
To reduce the exponent occurring in the second term, we apply integration
by parts, noting that
D(b−(m−1) ) =
−(m − 1) D(b)
.
bm
42
Symbolic integration
This yields
Z
−b−(m−1)
−v b−(m−1)
v · D(
)=
+
m−1
m−1
Z
D(v) ·
b−(m−1)
.
m−1
The integral of a/bm then reduces to
Z
a
−v
=
+
m
b
(m − 1)bm−1
Z
u + D(v)/(m − 1)
.
bm−1
So the exponent has been reduced by 1 at the cost of introducing an extra
rational function. By repeating this process we can get down to exponent 1
in the denominator.
4.3.7
To put the above ingredients in a systematic scheme, we need one more
item, viz., a suitable partial fraction expansion. The expansion we have in
mind is not the usual one from calculus. In calculus one uses the following
expansion to deal with a fraction a/b. Division with remainder reduces
to the case a/b = u + p/b with p = 0 or deg(p) < deg(b), and with u a
ms
1
polynomial. Expand b as product of irreducible factors: b = pm
1 · · · ps .
Then there exist polynomials aij each of which is either zero or satisfies
deg(aij ) < deg(pj ) such that
p X
=
b
j
Ã
am j
a1j
a2j
+ 2 + · · · + jj
pj
pj
pj
!
.
The bottle–neck in computations is the decomposition into irreducible factors. To avoid this, one uses a partial fraction expansion based on so–called
squarefree factorizations. This is more efficient in computations, see the exercises for an algorithmic approach. A polynomial b is called squarefree if
there exists no polynomial d of positive degree such that d2 divides b.
4.3.8 Lemma. (Squarefree factorization) Let b ∈ K[x] be a monic polynomial of positive degree. Then there exist relatively prime monic squarefree
polynomials b1 , b2 , . . . , bk with bk of positive degree such that
b = b1 b22 · · · bkk .
Proof. To produce such a factorization, factor b into irreducible factors and
regroup according to exponent.
¤
4.3 Rational functions
43
4.3.9 Example. A squarefree decomposition of x2 (1 + x2 )2 (1 + x + x2 )5 ∈ Q [x]
is
[x(1 + x2 )]2 · (1 + x + x2 )5 .
Squarefree decompositions are not unique in general. For example, x3 (1+x)6
admits the squarefree factorizations x3 (1 + x)6 and (x(1 + x)2 )3 .
4.3.10 The proof of the above lemma on the existence of squarefree factorizations
uses the decomposition into irreducible factors. There is, however, a more efficient way to obtain such a factorization which only uses gcd computations.
This approach is worked out in the exercises.
4.3.11 If the polynomial b (of positive degree) is represented as a product f1 · · · fm
of relatively prime polynomials fi , then every fraction a/b admits a partial
fraction decomposition of the form
a
a1
am
= a0 +
+ ··· +
,
b
f1
fm
where every ai (= 1, . . . , m) is either 0 or satisfies deg(ai ) < deg(fi ).
Similarly, every fraction a/bm with deg(a) < m deg(b) admits a partial
fraction decomposition of the form
a
a1 a2
am
=
+ 2 + ··· + m,
m
b
b
b
b
with each ai satisfying either ai = 0 or deg(ai ) < deg(b).
4.3.12 (Hermite’s method) Here is how we reduce the integral of a given rational expression to one in which only integrals of rational expressions with
squarefree denominators occur.
Let p, q ∈ K(x) be relatively prime and let q be monic. Hermite’s method
consists of reducing the integral of p/q to an expression of the form
Z
a
r
+
,
s
b
where a, b, r, s ∈ K[x], deg(a) < deg(b), gcd(a, b) = 1, b monic and squarefree.
• Using division we can write p = q u + v with v = 0 or deg(v) < deg(q).
This reduces the problem to the integration of v/q.
44
Symbolic integration
• Use the squarefree factorization
q = Πm
k=1 qk
to produce a partial fraction expansion of the form
m
k
u X X tkl
=
,
v
qkl
k=1 l=1
where deg(tkl ) < deg(qk ) if deg(qk ) > 0 and tkl = 0 otherwise.
By applying 4.3.6 to terms of the form
Z
tkl
qkl
with l > 1 repeatedly, reduce to the form asserted above.
4.3.13
Once an integral is reduced to integrals in which the denominators of the
rational expressions are squarefree, we can apply 4.3.5 to integrate at the
expense of having to introduce logarithmic extensions. To deal with these
integrals a bit more precise, we state the following result (for resultants, see
the appendix)
4.3.14 Theorem. (Rothstein–Trager) Let K(x) be a differential field in one
variable with field of constants K. Let a, b ∈ K[x] be relatively prime with
deg(a) < deg(b) and with b monic. Then the minimal algebraic extension of
K such that the integral of a/b can be expressed as
Z
a X
=
ci log(vi ),
b
i
with ci ∈ K and vi ∈ K[x], is the splitting field K of the resultant
R(z) = resx (a − z D(b), b) ∈ K[z].
In fact, the ci are the roots of R(z) and vi = resx (a − ci D(b), b) for every i.
Proof. First we determine the ci as roots of the resultant mentioned. We
start with the equality
a X vi0
=
ci
b
vi
i
4.4 Beyond rational functions
45
and clear denominators. In doing this, it is useful to write ui = Πj6=i vj :
a Πnj=1 vj = b
X
ci vi0 ui .
i
From this equality we deduce that b|v1 · · · vn . The converse also holds: since
vj |ui for j 6= i, the same equality implies that vj |bvj0 uj . But gcd(vj , vj0 ) = 1
since vj is squarefree and gcd(vj , uj ) = 1 by construction, and so vj |b. Again
using that the vi are relatively prime, we conclude that the product v1 · · · vn
divides b.
P
Using the equality b = v1 · · · vn , we find a = i ci vi0 ui . Then
a − cj b0 =
X
i
ci vi0 ui − cj
X
vi0 ui =
i
X
i
(ci − cj )vi0 ui .
Now vj divides each term in the last sum (for i = j the term vanishes!), so
that vj is a common divisor of a − cj b0 and b. But that implies that cj is a
root of res(a − zb0 , b).
In the next step we show that vi = gcd(a − ci b0 , b).
¤
4.4
Beyond rational functions
This section is intended to give an impression of the situation for transcendental expressions. Since a full treatment is beyond the scope of this course,
we will focus on special cases. In particular, we will concentrate on integrals
containing a logarithmic or exponential expression transcendental over the
field of rational functions, like
Z
Z
x
x2
or
e
,
log(x)
but will not touch upon integrals whose integrand is algebraic over this field,
like
Z
1
√
.
8
x +1
The third type seems closer to the rational functions than the first two types,
and one is therefore tempted to think that the third type is easier than the
other two. As it turns out, the first two types are simpler to deal with. A
treatment of the third type requires a working knowledge of the theory of
algebraic functions.
46
Symbolic integration
4.4.1 Definition. If L is a differential extension field of K, then L is called a
transcendental elementary extension of K if L is of the form
L = K(θ1 , . . . , θn ),
where each θi is logarithmic or exponential over K(θ1 , . . . , θn−1 ) and is transcendental (i.e., non–algebraic) over K(θ1 , . . . , θn−1 ). If there is no condition
on being trancendental, then we simply speak of an elementary extension of
K.
Usually, K will be Q (x).
4.4.2 Remark. Note that if α is transcendental over K, then K[α] ∼
= K[X] (the
∼
polynomial ring in one variable over K) and K(α) = K(X) under the map
sending X to α. This enables us to speak of the degree in α of a polynomial
expression in α.
4.4.3 Theorem. (Liouville’s principle) Suppose the integral of f ∈ K exists
in an elementary extension L of K, which has the same field of constants as
K. Then
Z
X
X v0
f = v0 +
ci log(vi ) or f = v00 +
ci i
vi
i
i
for some constants ci and elements vi ∈ K.
Proof. We will prove this in the special case where L = K(θ) with θ transcendental and logarithmic, θ 0 = u0 /u. The assumption implies that we can
write
Z
a(θ)
f=
,
b(θ)
with a and b relatively prime and with b monic. Again simplifying, we
assume that b(θ) = c(θ)m is the factorization into irreducible factors. Then
applying partial fraction expansion yields
m
X aj (θ)
a(θ)
= a0 (θ) +
b(θ)
c(θ)j
j=1
with deg(aj ) < deg(c(θ)). Diferentiating both sides gives us
0
f = a0 (θ) +
m ·
X
aj (θ)0
j=1
¸
jaj (θ)c(θ)0
−
.
c(θ)j
c(θ)j+1
4.4 Beyond rational functions
47
The left–hand side of this equality is independent of θ, so the right–hand
side must be independent of θ. If c(θ) is of positive degree, then c(θ)0 has
degree less than deg(c(θ)) and so is relatively prime with c(θ). But then
there is no term on the right–hand side to cancel
mam (θ)c(θ)0
.
c(θ)m+1
This contradiction implies that we get f = a0 (θ)0 . But then a0 (θ) must be
(at most) linear1 in θ, i.e., a0 (θ) = c θ + d with c a constant in K and d ∈ K.
In other words,
Z
f = d + c log(u).
The proof in the exponential case is similar, the proof in the case of a purely
algebraic extension is simpler. The general case needs some care.
¤
4.4.4 The following theorem provides a criterion for deciding when an integral is
elementary (i.e., exists in an elementary extension), given that the integrand
is a rational expression in a single transcendental logarithmic variable.
4.4.5 Theorem. Let K be a differential field with field of constants C and let
K(θ) be a transcendental logarithmic extension of K with the same field of
constants. Suppose p(θ)/q(θ) satisfies
a) gcd(p(θ), q(θ)) = 1;
b) deg p(θ) < deg q(θ);
c) q(θ) is monic and squarefree.
Then
Z
p(θ)
q(θ)
is elementary if and only if the zeros of R(z) = Resθ (p(θ) − zq(θ)0 , q(θ)) are
constants.
Proof. We restrict
R ourselves to the main aspects of the proof in a special case.
First suppose (p(θ)/q(θ)) is elementary. Liouville’s principle produces an
expression for p(θ)/q(θ). Assume for the sake of simplicity that this form is
p(θ)
v(θ)0
= v0 (θ)0 + c
,
q(θ)
v(θ)
1
This requires proof, for which we refer to the exercises
48
Symbolic integration
where we may assume that v(θ) ∈ K[θ] and is squarefree.
Consider the term v0 (θ) = a(θ)/b(θ) for some relatively prime a(θ), b(θ).
If the derivative v0 (θ)0 6= 0, it will contribute a square to the denominator
of p(θ)/q(θ) contradicting the assumption. So v0 (θ) = a(θ), a polynomial
expression in θ.
From the remaining equality
v(θ)0
p(θ)
=c
q(θ)
v(θ)
we deduce q(θ) = v(θ). But then p(θ) − cq(θ)0 = 0, so v(θ) = gcd(p(θ) −
cq(θ)0 , q(θ)) showing that c is a zero of the resultant R(z) = resθ (p(θ) −
zq(θ)0 , q(θ)).
If R(d) = 0, then it follows that deg gcd(p(θ) − dq(θ)0 , q(θ)) > 0. Let g
be an irreducible factor. Then g|q = v and g|p − dq 0 = cv 0 − dv 0 = (c − d)v 0 .
Since gcd(v, v 0 ) = 1, we conclude that c = d.
¤
4.4.6
Using our previous results we can refine the theorem in the case the integral
is elementary. Then
Z
m
p(θ) X
ci log(vi (θ)),
=
q(θ)
i=1
where the ci are the distinct roots of the resultant R(z) and where vi equals
gcd(p(θ)−ci q(θ)0 , q(θ)). In fact, K = K(c1 , . . . , cm ) is the minimal algebraic
extension of K such that the integral can be expressed in such a form.
4.4.7 Example. Consider the integral
Z
1
,
log(x)
i.e., the integrand is of the form above with p(θ) = 1 and q(θ) = θ, where
θ = log(x). So we work in the differential extension field Q (x, θ) of Q (x).
This is a transcendental logarithmic extension of Q (x). The resultant
R(z) = resθ (1 − z q(θ)0 , q(θ)) = resθ (1 −
z
z
, θ) = 1 −
x
x
has no constant zeros, so we conclude that the integral is not elementary.
4.4.8 Example. The integrand of
Z
1
x log(x)
4.4 Beyond rational functions
49
is of the form p(θ)/q(θ) with p(θ) = 1/x and q(θ) = θ. This time the
resultant is
1
z
1−z
R(z) = resθ ( − , θ) =
,
x x
x
with zero z = 1. This implies that the solution is 1 · log(v) with v =
gcd(p(θ) − q(θ)0 , q(θ)) = θ. So
Z
1
= log(log(x)).
x log(x)
R
4.4.9 Example. The theorem does not apply to the integral log(x2 + 1) since
the degree of the numerator (the numerator is θ = log(x2 + 1) of degree
1) is greater than the degree of the denominator. This example represents
another extreme of our problem, namely where the expression is polynomial
in θ.
4.4.10 Similar to the case of a single logarithmic extension is the case of a single
exponential extension.
4.4.11 Theorem. (Single exponential extension) Let K be a differential field
with field of constants C and let K(θ) be a transcendental exponential extension of K with the same field of constants. Suppose p(θ)/q(θ) satisfies
a) gcd(p(θ), q(θ)) = 1 and θ does not divide q(θ);
b) deg p(θ) < deg q(θ);
c) q(θ) is monic and squarefree.
Then
Z
p(θ)
q(θ)
is elementary if and only if the zeros of R(z) = Resθ (p(θ) − zq(θ)0 , q(θ)) are
constants.
4.4.12 Again, there is an explicit form if the integral is elementary. Let θ 0 /θ = u0 ,
then the explicit form is
X
X
−
ci deg(vi (θ))u +
ci log(vi ),
i
i
where the ci are the distinct roots of the resultant R(z), and vi = gcd(p(θ) −
ci q(θ)0 , q(θ)). The minimal finite extension of K over which such an explicit
form exists is K with the ci adjoined.
50
Symbolic integration
4.4.13 Example. This theorem covers the case
Z
1
,
2
x
e +1
R
2
but doesn’t cover the case e−x . In the first case we find R(z) = −1 − 2xz
with no constant zeros, so the integral is not elementary. For the integral
Z
x
,
ex2 + 1
the resultant R(z) = −2xz − x = x(−2z − 1) with zero z = −1/2. So this
integral is elementary. Using 4.4.12, this leads to the following expression
for the integral:
1 2 1
2
x − log(ex + 1),
2
2
since the gcd equals θ + 1.
4.4.14
The various results above describe integrals where the integrand is a quotient in which the degree of the numerator is less than the degree of the
denominator. In particular, this excludes polynomial expressions from the
range of application of these results! At first glance, they would seem the
easiest.
4.4.15 (Polynomial in an exponential) Here, we outline the procedure for elementary integrals where the integrand is a polynomial in eu and e−u . Let
K be a differential field with field of constants C, and let K(θ) be a transcendental exponential extension with θ 0 /θ = u0 , where u ∈ K. Consider a
Laurent polynomial
l
X
p(θ) =
pj θ j .
i=−k
Using Liouville’s principle one can reduce to the case that p(θ) is of the form

0
l
X v0
X
p(θ) = 
ci i ,
qj θ j  +
vi
j=−k
i
with the qj to be determined. Since (qj θj )0 = (qj0 + ju0 qj )θj , equating
coefficients in our two expressions for p(θ) yields the following system of
equations in the qi :
pj = qj0 + ju0 qj (j 6= 0)
P
p0 = (q0 + i ci log(vi ))0 .
4.4 Beyond rational functions
51
R
The second equation can be solved if and only if p0 is elementary. If this
is not the case, then the original integral is not elementary; otherwise, we
proceed to the equations with j 6= 0. These are of the form
y 0 + f y = g,
with f, g ∈ K. This differential equation is called the Risch differential
equation. We need to solve it withinR K. If any of these differential equations
fails to have a solution in K, then p(θ) is not elementary.
Solving the Risch equations brings us into the realm of symbolic differential equations, the topic of one of the projects for this course.
R 2
2
2
4.4.16 Example. Consider ex and let θ = ex ∈ Q (x, ex ). Here the situation
reduces to one Risch differential equation
1 = q10 + 2x q1 ,
for q1 ∈ Q (x). Substituting a rational expression in x for q1 easily leads to
an inconsistency demonstrating that the integral not elementary.
4.4.17 (Polynomial in a logarithm) There is a similar discussion for the integration of polynomial expressions in θ, where θ is logarithmic. The expression to integrate is of the form
p(θ) = pl θl + pl−1 θl−1 + · · · + p0 ,
where the pi ∈ K. If the integral is elementary then Liouville’s principle
(and a little arguing) shows that p is also of the form
p(θ) = v0 (θ)0 +
X
i
ci
vi0
vi
with v0 (θ) ∈ K[θ] and vi ∈ K. The ci are algebraic over K as usual.
From the two expressions we obtain that deg(v0 (θ)) = l + 1, i.e., v0 (θ) =
ql+1 θl+1 +· · ·+q0 . Equating both expressions yields a system of equations in
the qj . Depending on whether these equations admit a solution, the integral
is elementary. Here is an example of how this works out.
R
4.4.18 Example. The integral x log(x) is of the required form with p(θ) = xθ
(where θ = log(x)), i.e., of degree 1. So we start with
Z
x θ = q 2 θ 2 + q1 θ + q 0 ,
52
Symbolic integration
and obtain the system of equations
q20 = 0,
x = 2q2 θ0 + q10 ,
0 = q1 θ0 + q00 .
For the moment we ignore the first equation (it says that q2 is a constant).
Integrating the second equation gives 21 x2 +γ = 2q2 θ +q1 , with γ a constant,
so that q2 = 0 and q1 = 21 x2 +γ. Substituting this into the third equation we
2
0 = ( 12 x2 +γ)θ 0 +q00 , so that − x2 = γθ 0 +q00 . Integrating yields − x4 = γθ +q0 ,
2
so that γ = 0 and q0 = − x4 . We find
Z
x log(x) =
1 2
x2
x log(x) − .
2
4
4.4.19 Example. The integral
Z
log(x)
x
works out similarly. The integrand is θ/x, with θ = log(x). Set
Z
θ
= q2 θ 2 + q1 θ + q 0 .
x
Then we get the system
q20 = 0,
(2q2 θ + q1 )0 = 1/x,
q1 θ0 + q00 = 0.
(The second equation uses the fact that q2 is a constant, a fact following from
the first equation.) From the second equation we conclude that 2q2 θ + q1 =
θ+γ (where γ is a constant), and therefore q2 = 1/2 and q1 = γ. Substituting
this in the third equation yields q0 = −γθ + δ, where δ is a constant, so that
γ = 0 since q0 is does not depend on θ. So q0 is a constant and we get
Z
log(x)
1
1
= θ2 + constant = log(x)2 + constant.
x
2
2
Appendix A
Algebraic prerequisites
A.1
Groups
A.1.1 Definition. A group G is a nonempty set with a binary operation that
satisfies:
• associativity: g(hk) = (gh)k for all g, h, k ∈ G.
• existence of an identity element: there exists e ∈ G such that eg =
ge = g for all g ∈ G.
• existence of an inverse element: for every g ∈ G, there exists an inverse
element g −1 ∈ G such that gg −1 = g −1 g = e.
The number of elements in G is called the order of G and it is denoted |G|.
A subgroup H of G is a nonempty subset of G satisfying:
(a) H is closed under the binary operation: for all h, k in H, we have
hk ∈ H.
(b) H is closed under taking inverses: for all h ∈ H, we have h−1 ∈ H.
A subgroup is itself a group.
A.1.2 Example. Two important examples:
• the symmetric group Sn . This is the group of all bijections of the set
{1, . . . , n}. The binary operation is the composition of two bijections
and the identity element is the identity map.
53
54
Algebraic prerequisites
• The set of invertible n × n–matrices over a field K is a group with
respect to matrix multiplication; the identity matrix serves as identity
element. We denote this group by GL(n, K).
A.1.3 Definition. A group G is generated by g1 , . . . , gn ∈ G if any g ∈ G can
be written as a product of the gi ’s and gi−1 ’s. We denote this as G =
hg1 , . . . , gn i; the gi ’s are called generators of G. In the case that G is finite,
it suffices to impose the condition that every element of G can be written
as a product of the gi ’s.
A cyclic group G is a group generated by one of its elements. That
means that there exists an element g ∈ G such that every element of G
equals g n for some n ∈ Z.
A.1.4 Definition. Let G and H be two groups. A group homomorphism φ : G →
H is a function satisfying
φ(ab) = φ(a)φ(b) for all a, b ∈ G
If φ is injective, φ is said to be a monomorphism. If φ is surjective, φ is said
to be a epimorphism. If φ is bijective, φ is said to be an isomorphism.
A.2
Rings, ideals and quotient rings
A.2.1 (Rings) In these notes rings are commutative with unit 1. A ring is called
a domain (or integral domain) if ab = 0 implies a = 0 or b = 0 for all
a, b ∈ R. An element a ∈ R is called a unit (or invertible element) if there
exists an element b ∈ R with ab = 1. The set of units is a group with respect
to multiplication and is denoted by R∗ . An element a is called a zerodivisor
if there exists a nonzero b with ab = 0. A ring is a domain if and only if
it contains no nonzero zerodivisors. A ring is called a field if every element
6= 0 is a unit, i.e., if R∗ = R \ {0}. Since the sets of units and zerodivisors
are always disjoint, every field is a domain.
A morphism (of rings) from R to S is a map f : R → S such that
f (0R ) = 0S , f (1R ) = 1S , f (a + b) = f (a) + f (b) and f (ab) = f (a)f (b) for
all a, b ∈ R. A bijective morphism is called an isomorphism. The inverse of
an isomorphism is also an isomorphism. In this case the rings R and S are
called isomorphic; this is denoted by R ∼
= S.
A.2.2 (Ideals) An ideal I in the ring R is a nonempty subset such that
a) a, b ∈ I ⇒ a + b ∈ I;
A.2 Rings, ideals and quotient rings
55
b) a ∈ I and r ∈ R ⇒ ra ∈ I.
If S is a subset of R, then (S) denotes the ideal generated by S. It is the
smallest ideal in R containing I. An explicit description is
(S) = {r1 s1 + · · · + rm sm | r1 , . . . , rm ∈ R, s1 , . . . , sm ∈ I and m ∈ Z≥1 }.
If I and J are ideals, then I + J = {a + b | a ∈ I, b ∈ J} is also an ideal in
R, the sum of I and J. This notion generalizes to the sum of an arbitrary
number of ideals. The ideal I is a maximal ideal if I 6= R and if there
exists no ideal strictly between I and R, i.e., if J is an ideal in R satisfying
I ⊂ J ⊂ R, then J = I or J = R. The ideal I is called a prime ideal if
a b ∈ I ⇒ a ∈ I or b ∈ I.
Every maximal ideal is a prime ideal, but the converse does not hold in
general. For instance, (2) ⊂ Z [X] is prime, but not maximal.
If J is an ideal in the ring S and if f : R → S is a morphism, then
−1
f (J) is an ideal in R. For J = (0), this ideal is called the kernel of f and
denoted by ker(f ).
A.2.3 (Quotient rings) If I is an ideal in R, then the quotient ring R/I is the
ring whose elements are the equivalence classes with respect to the equivalence relation
a ∼ b ⇔ a − b ∈ I.
The equivalence class containing a is denoted by a+I, a or just a if confusion
is not likely to occur. The ring operations on R/I are
a) (addition) (a+I) + (b+I)=(a+b)+I;
b) (multiplication) (a+I) (b+I)=(ab)+I.
The zero element is 0 + I and the unit element 1 + I. The map
p : R → R/I,
p(a) = a + I
is a morphism and often called the canonical map (or natural map) from R
to R/I. The kernel of p is I.
A.2.4 Theorem. Let I be an ideal in the ring R.
a) I is a prime ideal if and only if R/I is a domain;
56
Algebraic prerequisites
b) I is a maximal ideal if and only if R/I is a field.
A.2.5 Theorem. (First Isomorphism Theorem) If f : R → S is a morphism
then f induces a unique injective morphism f : R/ ker(f ) → S such that
f (a) = f (a) for every a ∈ R. If f is surjective, then the induced morphism
f is an isomorphism.
A.2.6
If J is an ideal in R then the ideals of R/J are in one–to–one corespondence
with the ideals in R containing J under the correspondence I ↔ p(I), where
p : R → R/J is the canonical map. The image p(I) of an ideal I containing J
is often denoted by I/J. The kernel of the composition of the natural maps
R → R/J → (R/J)/(I/J) is equal to I. Applying the First Isomorphism
Theorem A.2.5 yields
A.2.7 Theorem. (Second Isomorphism Theorem) If J ⊂ I are ideals in R,
then
R/I ∼
= (R/J)/(I/J).
A.3
Finite fields
For p prime, the ring Z/pZ is a finite field. But there are many more finite
fields.
If F is a field the kernel of the morphism
Z → F, m 7→ sign(m)(1 + · · · + 1) (with |m| terms 1) if m 6= 0 and 0 7→ 0
has kernel (0) or pZ for some prime p. If F is a finite field only the latter
case can occur and F is said to have characteristic p. (If for a field the
kernel of the map is (0) the field is said to have characteristic 0.) The First
Isomorphism Theorem A.2.5 shows that a finite field contains a copy of the
field Z/pZ with p elements. In particular, the field F is a finite dimensional
vector space over Z/pZ. If the dimension is m then F has pm elements. So
the number of elements of a finite field is necessarily a prime power.
A.3.1 Theorem. Let F be a field with pm elements where p is a prime.
m
a) Every element a ∈ F satisfies ap = a.
b) The group of units F ∗ is cyclic of order pm − 1.
c) If L is also a field with pm elements then F and L are isomorphic.
A.4 Resultants
57
d) There exists an irreducible polynomial f ∈ (Z/pZ )[X] such that F ∼
=
(Z/pZ )[X]/(f ).
A.3.2 Since finite fields with the same number of elements are isomorphic, one
uses a single notation for a field with q (with q a prime power) elements: Fq .
A.4
Resultants
A.4.1 Let f, g ∈ k[X] be two polynomials of positive degree. The resultant of f, g
is an element of k that determines if f and g have a common factor or not:
f and g have a nonconstant common factor if and only if the resultant is 0.
If f and g have a common factor h, then there is a polynomial relation of
the form Af + Bg = 0 with deg(A) < deg(g) and deg(B) < deg(f ): simply
take A = g/h and B = −f /h. The following lemma states that the converse
also holds.
A.4.2 Lemma. Let f and g be of positive degree. Then f and g have a factor in
common if and only if there exist nontrivial polynomials A and B satisfying
deg(A) < deg(g) and deg(B) < deg(f ) such that
Af + Bg = 0.
Proof. One implication was shown above, so we assume that we have a
relation Af + Bg = 0 as in the statement of the lemma and that f and g are
relatively prime. From the last assumption, we deduce from the euclidean
algorithm that there exists a relation uf + vg = 1. Multiplying this relation
by B we obtain uf B+vgB = B. Using Bg = −Af we find B = uf B−vAf =
(uB − vA)f contradicting the assumption deg(B) < deg(f ).
¤
A.4.3 To check the existence of A and B comes down to solving a system of linear
equations. If
f = f0 +f1 X+· · ·+fm X m
(f0 6= 0, fm 6= 0)
and
g = g0 +g1 X+· · ·+gn X n
and if we write
A = a0 + a1 X + · · · + an−1 X n−1 ,
B = b0 + b1 X + · · · + bm−1 X m−1 ,
(g0 6= 0, gn 6= 0)
58
Algebraic prerequisites
then substitution in the equation Af + Bg = 0 gives a system of linear
equations, whose coefficient matrix is the transpose of the (m + n)–matrix


f0 f1 · · ·
fm


f0 · · · fm−1 fm




..


.




f
f
·
·
·
f
0
1
m

A(f, g) = 
 g0 g1 · · ·

gn




g0 · · ·
gn




..


.
g0
· · · gn
This system has a nontrivial solution if and only if the determinant det(A(f, g)) =
0.
Note that if the resultant is 0, the gcd of f and g is a nontrivial common
factor.
A.4.4 Definition. The determinant of the matrix in (A.4.3) is called the resultant of f and g and denoted by R(f, g). The notation R(f, g, x) is used to
emphasise that we view f and g as polynomials in x.
A.4.5 Remark. Although not stated explicitly, the resultant is also usefull for
polynomials in several variables. For instance, the polynomials X 3 + XY 2 +
Y X 2 + Y 3 and X + Y can be viewed as polynomials in x with coefficients
in Q (y): Y 3 + (Y 2 ) X + Y X 2 + X 3 and Y + X. The resultant is
¯ 3
¯
2 Y
¯ Y
¯
Y
1
¯
¯
¯ Y
1 0 0 ¯¯
¯
= 0,
¯ 0 Y
1 0 ¯¯
¯
¯ 0
0 Y 1 ¯
showing that the two polynomials have a common factor in Q (Y )[X]. By
clearing denominators, a common factor in Q [X, Y ].
A.4.6 Example. Let f = ax2 + bx + c with a 6= 0. Then
¯
¯
¯ c b a ¯
¯
¯
R(f, f 0 ) = ¯¯ b 2a 0 ¯¯ = −a (b2 − 4ac).
¯ 0 b 2a ¯
In this expression, the well-known discriminant b2 − 4ac of a quadratic polynomial occurs. We regain the result that f has a zero of multiplicity 2 if
and only if b2 − 4ac = 0.
A.5 Groups
59
A.4.7 Proposition. Let f, g ∈ k[X] be polynomials of positive degree.
a) f and g have a common factor if and only if R(f, g) = 0.
b) R(f, g) ∈ (f, g), i.e., there exist polynomials u and v such that uf +
vg = R(f, g).
Proof. The first item was shown above, so we turn to b). If R(f, g) = 0,
then we can take u = v = 0. If R(f, g) 6= 0, then f and g are relatively
prime, so there exist ũ and ṽ with ũf + ṽg = 1. Multiplying through by
R(f, g) gives the required identity.
¤
A.4.8 For computations, the determinant in the definition of the resultant is quite
inefficient. In the exercises a more efficient way to compute the resultant is
described.
A.5
Groups
One of the most important notions in Mathematics is the notion of a group:
A.5.1 Definition. A group G is a non empty set with a binary operation that
satisfies:
• associativity: g(hk) = (gh)k for all g, h, k ∈ G.
• existence of the identity element: there exists e ∈ G such that eg =
ge = g for all g ∈ G.
• existence of an inverse element: for every g ∈ G, there exists an inverse
element g −1 ∈ G such that gg −1 = g −1 g = e.
The number of elements in G is called the order of G and it is denoted
|G|.
A.5.2 Example. The integers Z, the rationals Q and the real R are groups with
respect to the addition + and identity element 0. However, Z is not a group
with respect to the multiplication.
A.5.3 Example. A very important example of group is the symmetric group Sn .
This is the group of all bijections of the set {1, . . . , n}. The binary operation
is the composition of two bijections and the identity element is the identity
map.
60
Algebraic prerequisites
A.5.4 Example. The set of invertible matrices over a field K is a group denoted
by GL(n, K).
A.5.5 Definition. A finite group G is generated by g1 , . . . , gn ∈ G if any g ∈ G
can be written as a product of the gi ’s. We denote G = hg1 , . . . , gn i and the
gi ’s are called the generators of G.
A cyclic group G is a group generated by one of its elements. That
means there exists an element g ∈ G such that G = {e, g, g 2 , . . . , g n } where
e denotes the identity element. We can also write G = hgi.
A.5.6 Definition. Let G and H be two groups. A group homomorphism φ : G →
H is a function satisfying
φ(ab) = φ(a)φ(b) for all a, b ∈ G
If φ is injective, φ is said to be a monomorphism.If φ is surjective, φ is said
to be a epimorphism. If φ is bijective, φ is said to be a isomorphism.
A.5.7 Example. Consider the group homomorphism φ : Sn → GL(n, K) defined
by: to each element σ ∈ Sn we associate the matrix φ(σ) = (aij )ni,j=1 with
aij =
(
1 if σ(j) = i,
0 otherwise
It is easy to see that Sn and φ(Sn ) are isomorphic. Using this representation
of Sn we can say that Sn is a subset of GL(n, K).
For instance, the permutation σ ∈ S3 with σ(1) = 2, σ(2) = 1, σ(3) = 3
has matrix:


0 1 0
1 0 0
0 0 1
A.5.8 Definition. A subgroup H of G is a non-empty subset of G satisfying:
(a) H is closed under the binary operation: for all h, k in H, we have
hk ∈ H.
(b) H is closed under taking inverses: for all h ∈ H, we have h−1 ∈ H.
A.5.9 Example. For every m ∈ Z, the subset mZ = {mn | n ∈ Z} is a subgroup
of the group Z.
A.5 Groups
61
A.5.10 Example. Consider the matrix
¶
µ
0 −1
A=
1 0
The set {I, A, A2 , A3 } is a subgroup of GL(2, C). Besides, this is a cyclic
group generated by A.
A.5.11 Example. An example of a subgroup we will use later is a finite matrix
group. It is defined as being a non-empty finite subset of GL(n, C) which is
closed under matrix multiplication. In the exercises, you will show that this
is actually a subgroup of GL(n, C). The example A.5.10 is a finite matrix
group. The symmetric group Sn seen as permutation matrices is also a finite
matrix group.
A.5.12 Example. The set of det 1 matrices of order n over a field K is a subgroup
of GL(n, K) denoted by SL(n, K).
62
Algebraic prerequisites
Bibliography
[1] W. Adams, P. Loustaunau (1994). An introduction to Gröbner bases.
Graduate Studies in Mathematics 3, AMS, Providence.
[2] M. Bronstein (1997). Symbolic integration I. Algorithms and Computation in Mathematics, Vol. 1, Springer, Berlin etc.
[3] B. Buchberger, F. Winkler (eds.) (1998) Gröbner Bases and Applications, London Mathematical Society Lecture Notes 251, Cambridge University Press.
[4] A.M. Cohen, H. Cuypers, H. Sterk (eds.) (1999). Some tapas of computer
algebra, Springer-Verlag, Berlin etc.
[5] D. Cox, J. Little, D. O’Shea (1992, 1997). Ideals, varieties and algorithms: an introduction to computational algebraic geometry and commutative algebra. Undergraduate Texts in Mathematics, Springer Verlag,
Berlin etc.
[6] K.O. Geddes, S.R. Czapor, G. Labahn (1992). Algorithms for computer
algebra. Kluwer Academic Publishers, Boston/Dordrecht/London.
[7] B. Sturmfels (1993) Algorithms in Invariant Theory, Text and Monographs in Symbolic Computation, Springer-Verlag, Wien.
63
Index
ideal membership test, 8
integration by parts, 40
S–polynomial, 9
algebraic number, 19
Landau–Mignotte bound, 33
leading coefficient, 5
leading monomial, 5
leading term, 5
least common multiple, 9
lexicographic order, 4
Liouville’s principle, 46
logarithm, 38
Berlekamp’s algorithm, 29, 31
Buchberger’s algorithm, 11
Chinese remainder theorem, 30
content, 27
degree of field extension, 20
derivation, 37
Dickson’s lemma, 6
differential extension field, 38
differential field, 37
differential operator, 37
minimal polynomial, 20
monomial, 2
(multi)degree, 2
total degree, 2
monomial ideal, 6
monomial ordering, 4
multidegree, 5
Eisenstein’s criterion, 28
elementary extension, 46
elimination, 13
exponential, 38
noetherian ring, 2
partial order, 3
Gröbner basis, 7
minimal, 9
reduced, 9
graded lexicographic order, 4
graded reverse lexicographic order,
4
Rothstein–Trager theorem, 44
squarefree factorization, 42
squarefree polynomial, 42
total order, 4
transcendental elementary extension, 46
Hensel lift, 32, 33
Hermite’s method, 43
Hilbert basis theorem, 2
Hilbert Nullstellensatz, 18
Weak, 18
well–ordering, 4
zeroset of an ideal, 3
64